The simplest thing to do is simply run the problem 'the other way' - can you show that if $x_n < 4$, then $x_{n+1} < 4$? Once you have that, induction will let you go 'from one to infinity'; since the statement is true for $x_1$, and since you've shown that if it's true for $x_n$ then it's true for $x_{n+1}$, you're then allowed to conclude by induction that it's true for all natural numbers $n$.
As for the so-called 'induction step' - showing that if the inequality is true for $x_n$, it's true for $x_{n+1}$ - you should be able to do that yourself, using just what you know about inequalities (a more specific hint: if $a < b$, then $(a+c) < (b+c)$; in addition, if $c > 0$, then $a\times c < b\times c$. These should be all that you need.)
Edit: Yes, the approach that you offer in your addendum is exactly what you're after! Nicely done.
The way you've described induction is a bit muddled, and I blame the fact that proofs using induction are often written very informally.
The conceptually simplest form of the principle of mathematical induction is this:
Let $S$ be a set of natural numbers with the following two properties:
$0\in S$
For all $n\in \Bbb N$, if $n\in S$ then $n+1 \in S$.
Then $S = \Bbb N$.
This principle can be recast in logical terms (as a "schema" as follows):
Let $P$ be a propositional function in $\Bbb N$. That is, for each $n\in \Bbb N$, let $P(n)$ be a statement.
Suppose that the following hold:
$P(0)$ is true.
For each $n\in \Bbb N$, if $P(n)$ is true then $P(n+1)$ is true.
Then we can conclude that for each $n\in \Bbb N$, $P(n)$ is true.
In fact, it's easy to see that either of these approaches can be extended to allow any initial value, not just $0$, at the cost of having the proposition hold (or the number be an element of the set) only for numbers greater than some value (in your case $1$).
Using one of these somewhat more formal approaches to induction should help you keep a clear idea in your mind of how you need to approach the proof.
Let $P(n)$ be the statement that $$f^{(n)}(x) = (-1)^{n-1}\frac{(n-1)!}{(1+x)^n}$$. Prove first that $P(1)$ is true. Then prove that for each $n\in\Bbb N$, if $P(n)$ is true then so is $P(n+1)$.
Best Answer
It works in the first case.
If it works in the first case, then it works in the second case.
If it works in the second case, then it works in the third case.
If it works in the third case, then it works in the fourth case.
$\ldots$ and so on.
That's what mathematical induction does. But all of the statements above beginning with "if" are proved in one argument, rather than one by one.