Compute limit by induction ("inductive step of existence" issue)

I see nothing wrong with your approach. You proved correctly that:

1. The limit $L(n)$ exists if and only if the limit $L(n+1)$ exists.
2. If both of them exist, then $L(n+1)=2L(n)+(n+1)2^n$.

So, now, since $L(1)$ exists, it's only a matter of using induction to compute the value of $L(n)$ for each $n\in\mathbb N$.

Your ideas are correct.In the second step i would suggest not to take limits because you do not know of the limit in the left hand side exists.Just do the computations of the limits on the right hand side of the equality which we know that exist from the induction hypothesis and etc.

Then after you compute these then you take limits to the left hand side.

using the fact that the limit of $f+gh$ at a point $x_0$ is equal to $\lim f+(\lim g \lim h)$

if $\lim g,\lim f, \lim h$ exist at $x_0$(and are real numbers).

Here it is another proof of the existence with the notion of derivative:

Also to make clear i use just the definition of derivative and not the L'Hospital rule because of the tag.

This limit exists $\forall n\in \mathbb{N}$ because:

Let $n \in \mathbb{N}.$

Take the function $f(x)=\prod_{k=1}^n(1+x^k)$ which is differentiable.

We have that $$\lim_{x \to 1}\frac{\prod_{k=1}^n(1+x^k) -2^n}{x-1}=\lim_{x \to 1}\frac{f(x)-f(1)}{x-1}=f'(1)$$

Now if you want, you can use induction to compute $f'(1)$ as an exercise.

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.