I know this should be obvious but somehow I can't seem to figure it out and it annoys me!
My definition of nilpotent groups is the following:
A group $G$ is nilpotent if every subgroup of $G$ is subnormal in $G$, or equivalently if $U<N_G(U)$ for all $U<G$. And my definition of solvable groups is that a group $G$ is solvable if $U'\neq U$ for all subgroups $1\neq U \le G$, where $U'$ is the commutator subgroup of $U$.
My question is then: why are nilpotent groups solvable?
[Math] Nilpotent groups are solvable
group-theorynilpotent-groupssolvable-groups
Related Solutions
This is a kind of long answer to explain why normal subgroups are so helpful for finding subgroups. Once you learn about composition series and chief series most of this is pretty easy. Derived series and lower central series make for easier homework questions, but if you don't see where they come from, then it might be comforting to know they are just one of many series that work out. I give some pretty casual series below that give equivalent definitions in finite groups.
The dihedral group of order 8 has subgroups of orders 1, 2, and 4. In fact it has normal subgroups of each of these orders, and every subgroup of order 4 has a subgroup of order 1 and 2.
In fact the same is true for any group of order 8. Something similar is true in any group of order $p^n$: it has normal subgroups of order $p^k$ for all $0 \leq k \leq n$ and every subgroup of order $p^k$ contains subgroups of order $p^i$ for all $o \leq i \leq k$.
The normal subgroups feature is particularly helpful since it allows us to replace the group of order $p^n$ by groups of order $p^k$ and $p^{n-k}$. That is, we can replace $G$ by $N$ and $G/N$ where $N$ is a normal subgroup. Subgroups of $N$ are clearly subgroups of $G$ (with small order), and subgroups of $G/N$ are all of the form $H/N$ where the $H$ are subgroups of $G$ (with large order).
So normal subgroups are nice as far as finding more subgroups in smaller groups. A normal series is just a way to break down a group into lots of small pieces. By normal series of length $d$, I'll mean subgroups $N_i$ such that $1=N_d$, $N_0 = G$, each $N_i$ is normal in $G$, and $N_i \leq N_{i-1}$. It turns out that studying how nice the subgroups of $G$ are is very related to how nice the subgroups between $N_i$ and $H=N_{i-1}$ are, when $N_i \leq N_{i-1} \leq G$.
Super
Here is an example: $N \leq H \leq G$ is super-nice if every subgroup between $N$ and $H$ is normal in $G$. Groups with a super-nice normal series are called supersolvable. They have subgroups of every possible order, and in fact every maximal chain of subgroups between any two subgroups has the same length [and these are the only finite groups with that property].
Proving that super-nice groups have all the subgroups of (ii) should be an easy exercise. Proving that the stronger version of (ii) [ that all maximal chains of subgroups have the same length ] is a theorem of Iwasawa. A related theorem of Huppert is that a group in which every maximal subgroup has index a prime is supersolvable (so if it has the “right” large subgroups, then it has all possible subgroups).
Duper
Doing better than super-nice is pretty hard, but it helps to know that being super nice means that $G$ acts like numbers when it acts on $H/N$. $N \leq H$ is super-nice means that $g^{-1}hg = h^k n$ for some number $k$ and some $n \in N$ (and with some mild hypotheses, the $k$ does not depend on $h$). What is the easiest choice of $k$? Well $k=1$ of course.
$N \leq H \leq G$ is super-duper-nice if for every $g\in G$ and $h\in H$, there is some $n \in N$ so that $g^{-1} h g = h n$. Notice $h^k$ is just plain old $h$ here. That's the duper.
A group of order $p^n$ has a super-duper-nice normal series (in fact a super-nice normal series is automatically super-duper-nice because the possible $k$ have to divide $[G:N]$ and be relatively prime to $[H:N]$, but they are both powers of $p$, leaving $k=1$ as the only possibility). A group that is a direct product of (its Sylow) $p$-groups has a super-duper-nice normal series. It turns out that a finite group with a super-duper-nice normal series is also a direct product of its Sylow $p$-subgroups, mostly because direct products tend to have $k=1$. Such groups are called nilpotent.
The exercises here should be the same as in your textbook.
Barely
$N \leq H \leq G$ is barely-nice if every subgroup between $N$ and $H$ is normal in $H$ (rather than $G$). This is still enough to get lots of subgroups but some are “missing:” if $G$ is a product of three primes, like $2 \cdot 2 \cdot 3$ then you can get a chain of subgroups of the right length, like order $1 < 2 < 2\cdot 2 < 2\cdot 2 \cdot 3$, but maybe you can't get other chains of the right length, $1 < 2 < 2\cdot 3 < 2 \cdot 3 \cdot 2$ is impossible in $A_4$. Sylow theorems help out to let you get as many primes as you want all in a row, but once you change primes you can't necessarily change back. A group is called solvable if it has a barely-nice normal series.
I think the exercises here are a bit more difficult. You should be able to prove that you can get a chain of subgroups in the form of (iii) without much trouble [ using one more prime power in each larger subgroup ], but proving you can choose the order of the primes is a little harder (1928 result of Hall). Proving that if you can do this the group is solvable is another result of Hall, but quite a bit harder.
Naughty
A group is said to be insoluble (or naughty) if it doesn't even have a barely-nice normal series. These groups are crazy and missing tons of subgroups. That's good in a way, since too many subgroups means there is too much to look at.
All finite groups have their subgroups of prime order. If $p$ divides $|G|$, then $G$ has a subgroup of order $p$. So small subgroups just tend to exist. It's the big ones that can vanish. If $G$ is even barely-nice and $p$ divides $|G|$, then $G$ has a subgroup of order $|G|/p^k$ for some $k$. For $G$ super-nice, $G$ has a subgroup of order $|G|/p$.
However, the group $S_{23}$ is pretty messed up. For $p=2$ and $p=23$ it has subgroups of order $|G|/p$. However, none of the other primes $p$ that divide its order (to be clear, that is $p=3,5,7,11,13,17,19$, a LOT of primes) even have subgroups of order $|G|/p^k$ for any positive integer $k$. In fact, other than $|G|/2$ and $|G|/23$ the next biggest subgroup has order $|G|/253$ and then $|G|/1771$. Not $|G|/3$, not $|G|/15$, not even a measly $|G|/200$.
In some sense this is no surprise, $S_{23}$ is almost simple, it has only one proper non-identity normal subgroup $A_{23}$. No normal subgroups means no real reason to have other subgroups. Well Sylow subgroups, and their normalizers. A few exceptional subgroups creep by, but the huge variety of subgroups forced into existence in the super-duper nice groups just isn't here.
In addition to what has been answered: a subgroup $H$ of $G$ is called subnormal if there exists a series of subgroups $H=H_0 \lhd H_1 \lhd \cdots \lhd H_s=G$. A normal subgroup is obviously subnormal, but the converse is not true. Now, finite nilpotent groups are exaclty the finite groups in which every subgroup is subnormal. For a proof: see for example M.I. Isaacs, Finite Group Theory, Lemma 2.1.
Best Answer
Let $G$ be nilpotent and nontrivial. Since every maximal subgroup of $G$ must be normal, and if $M$ is maximal then $G/M$ has no proper subgroups, it follows that if $M$ is maximal then $G/M$ is a group of order $p$, hence abelian. Therefore, $[G,G]\subseteq M$, since $[G,G]$ is contained in any normal subgroup $N$ of $G$ such that $G/N$ is abelian. In particular, $[G,G]\neq G$. Now simply note that being nilpotent is inherited to subgroups, as proven below, to conclude that $[H,H]\neq H$ for all subgroups $H$ of $G$ when $G$ is nilpotent. Hence, if every subgroup of $G$ is subnormal ($G$ is nilpotent), then the commutator subgroup of $H$ is properly contained in $H$ for any nontrivial subgroup $H$ of $G$ ($G$ is solvable).
(If $H\leq G$ and $K$ is a subgroup of $H$, then $K$ is subnormal in $G$, so there exist subgroups $K\triangleleft K_1\triangleleft K_2\triangleleft\cdots\triangleleft K_m=G$. Intersecting the subnormal series with $H$ gives you a subnormal series fo $K$ in $H$, showing $K$ is subnormal in $H$ as well; thus, every subgroup of $H$ is subnormal, so subgroup of nilpotent is nilpotent).
Added. I am tacitly assuming above that $G$ has maximal subgroups; so it might fail for infinite groups in which every subgroup is subnormal. In the infinite case, the usual definition of "nilpotent" is via either the upper central series or the lower central series, and that of "solvable" via the derived series. In the case of the definition via the lower central series, proving solvability is very easy: recall that the lower central series of $G$ is defined inductively by letting $G_1=G$ and $G_{n+1}=[G_n,G]$; and a group $G$ is nilpotent if and only if $G_{n+1}=\{1\}$ for some $n\geq 1$. Now note that $G^{(2)}=[G,G]=G_2$, and if $G^{(k)}\subseteq G_n$, then $G^{(k+1)} = [G^{(k)},G^{(k)}] \subseteq [G_k,G]=G_{k+1}$. So if the lower central series terminates, then so does the derived series, proving that if $G$ is nilpotent then $G$ is solvable.