Proof for Generating function of Bessel function I have already seen the relative questions

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*Generating function for Bessel function

*Prove the generating function for the Bessel function.
but I need some help for the reasoning behind a specific step in the proof the title suggests.

I believe that in the highlighted step we are adding more terms than we should. For instance if $$n = -100$$ in the outer sum, then in the inner sum the first term should not be $$k = 0$$ because then $$m = k + n = -100 < 0$$
Also if you could point out a resource with a more comprehensive approach to the proof, I would appreciate it. Thank you in advance!
Link to proof(page 9): https://pdxscholar.library.pdx.edu/cgi/viewcontent.cgi?article=1745&context=honorstheses
 A: Ok, so problem solved eventually...
Consider this product:
$$P = \sum_{m = 0}^{\infty} \frac{(xz/2)^m}{m!} \sum_{k = 0}^{\infty} \frac{(-x/(2z))^k}{k!} = \sum_{k = 0}^{\infty}\sum_{m = 0}^{\infty} (-1)^k \frac{(x/2)^{m+k}}{m!k!} z^{m-k} $$
and substitute m-k with n, then
$$P = \sum_{k = 0}^{\infty}\sum_{n = -k}^{\infty} (-1)^k \frac{(x/2)^{n+2k}}{(n+k)!k!} z^{n} = \sum_{k = 0}^{\infty}\sum_{n = -k}^{\infty} (-1)^k \frac{(x/2)^{n+2k}}{\Gamma(n+k+1)k!} z^{n}$$
Now the tricky part left to do is recognize that $\Gamma(l) = \infty , \forall l \in \{0,-1,-2,-3,...\}$.
Therefore $\frac{1}{\Gamma(l)} = 0$,
which means that the index n can now take also values like $\{-k-1,-k-2, ...\}$ down to $-\infty$.
So we have:
$$P = \sum_{k = 0}^{\infty}\sum_{n = -\infty}^{\infty} (-1)^k \frac{(x/2)^{n+2k}}{\Gamma(n+k+1)k!} z^{n}$$
and by reordering the sum(if it is possible?) we obtain the desired generating function to be equal to
$$P = \sum_{n = -\infty}^{\infty}\sum_{k = 0}^{\infty} (-1)^k \frac{(x/2)^{n+2k}}{\Gamma(n+k+1)k!} z^{n} = \sum_{n = -\infty}^{\infty}J_n(x) z^n$$
