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.