TITLE: Compute limit by induction ("inductive step of existence" issue)
QUESTION [7 upvotes]: I was asked to compute the limit 
$\lim\limits_{x \to 1} \dfrac{\displaystyle\prod_{k=1}^n (1+x^k) - 2^n}{x-1}$
My approach was as follows 
i) Define $L(n):=\lim\limits_{x \to 1} \dfrac{\displaystyle\prod_{k=1}^n (1+x^k) - 2^n}{x-1}$
ii) $L(1)$ exists and trying to write $L(n+1)$ in terms of $L(n)$, it turns out that $L(n+1)$ exists iff and $L(n)$ exists (from second line of the following equalities)
$$\begin{align*} L(n+1) & = \lim\limits_{x \to 1} \dfrac{\displaystyle\prod_{k=1}^{n+1} (1+x^k) - 2^{n+1}}{x-1} \\ & = \lim\limits_{x \to 1} \left( 2\times \dfrac{\displaystyle\prod_{k=1}^n (1+x^k) - 2^n}{x-1} + \lim\limits_{x \to 1} \dfrac{x^{n+1}-1}{x-1}\times \prod_{k=1}^n (1+x^k)\right) \\ & = 2L(n) + (n+1)\times 2^n \end{align*}$$
iii) the algebraïc identity linking $L(n)$ and $L(n+1)$ is $L(n+1) = 2L(n) + (n+1)\times2^n$
iv) Then I defined $l(n):=\dfrac{L(n)}{2^n}$ noticing that it satisfies the telescopable relation $l(n+1)-l(n) = \dfrac{n+1}{2}$
v) Computing $l(n)$ then $L(n)$ now is easy and it yields $L(n) = 2^{n-1}\times \dfrac{n(n+1)}{2}$
My question is that a friend objected step (ii) (third line in aligned equations) telling me that the problem of existence is not solved yet the way I did it, and that we can't do such a manipulation when expressions we manipulate are not justified beforehand to exist.  As I see it, since $L(1)$ exists and figuring out that $L(n+1)$ exists iff $L(n)$ does, is enough to this issue of existence.
I wonder whether I'm right or wrong and is the proof I provided rigorous enough, otherwise what do I need it to add to get it better
Thanks in advance for any suggestions / advice.

REPLY [4 votes]: You did right: the proof of existence of $L(n)$, for all $n\ge1$, is good because you know that $L(1)$ exists and that $L(n+1)$ exists as soon as $L(n)$ does.
A different strategy showing the power of differential calculus is to realize that this is the derivative at $1$ of
$$
f_n(x)=\prod_{k=1}^n(1+x^k)
$$
and that, for $x>0$,
$$
\log f_n(x)=\sum_{k=1}^n \log(1+x^k)
$$
Hence
$$
\frac{f_n'(x)}{f_n(x)}=\sum_{k=1}^n\frac{kx^{k-1}}{1+x^k}
$$
and, taking into account that $f_n(1)=2^n$, we get
$$
f_n'(1)=2^n\sum_{k=1}^n\frac{k}{2}=2^{n-2}n(n+1)
$$
No fancy telescoping and other troubles. I understand you have not yet done derivatives; when you'll have done them, check back and judge for yourself what's the best strategy.