Remember that we're dealing with a partial order! This is fundamentally different from a total order like $\leq$ is on numbers.
Because: For all numbers, we have either $a\leq b$ or $b\leq a$, i.e. everything is comparable.
However a partial order is partial because elements need not be comparable. Solely if elements should happen to be comparable, then the partial order imposes some laws like transitivity.
Now take as an example the proper subsets of $\{1,2,3\}$ with the partial order of $\subseteq$. E.g. $\{1\} \subseteq \{1,2\}$, but the sets $\{1,2\}$ and $\{2,3\}$ simply are not comparable because neither includes the other. However there are no "bigger" proper subsets to each of them, so both are maximal.
Now if in turn all elements were comparable, the maximum was unique (hence in total orders it is). But in partial orders, this isn't usually the case.
If you haven't already, you should prove that for all integers $k>0$, if you have an integer $N>0$ such that $2^{N}>N^{k}$, then, for all $n\geq N$, $2^{n}>n^{k}$ (or, more generally, there is an integer $N$ such that $2^{n}>p\left(n\right)$ for all $n\geq N$ where $p$ is any polynomial). This is a simple induction argument, and should be done in any algorithms course. From there, you set up the problem like so:
You're given that $8n^{2}<64n\lg n$. So, after a bit of algebra (you can do this part), we get it in to the form $2^{n}<n^{8}$. We have to find all $n>0$ such that this holds, and so, once we have found $N$ such that $2^{N}<N^{8}$ and $2^{N+1}\geq\left(N+1\right)^{8}$, we will know that all integers in the interval $1\leq n\leq N$ satisfy the problem. You can easily find $N$ with a calculator.
(Hint: $N$ is bigger than 40 but less than 45)
Best Answer
Hint: $$\sum_{k=0}^{\infty}k^2x^k= x\sum_{k=0}^{\infty}k^2x^{k-1}= x\frac{d}{dx}\sum_{k=0}^{\infty}kx^k$$
Can you follow on from this?
Edit: some more detail:
The idea with a question like this is to transform the question into something we already know. In this case, the sum looks remarkably similar to $$\sum_{k=0}^\infty x^k = \frac{1}{1-x} \text{ for } 0<x<1$$
Can you see a way to transform (perhaps by differentiating) the original sum into the simpler one above?
Continuation (if you get stuck):
$$x\frac{d}{dx}\sum_{k=0}^{\infty}kx^k = x\frac{d}{dx}\left( x\sum_{k=0}^{\infty}kx^{k-1}\right)= x\frac{d}{dx}\left( x\frac{d}{dx}\sum_{k=0}^{\infty}x^k\right)= x\frac{d}{dx}\left( x\frac{d}{dx}\left(\frac{1}{1-x}\right)\right)$$