Limit,Sum,Fraction(2)

Differentiating the standard geometric series $n$ times and dividing by $n!$ on both sides gives the identity $\sum_{k=0}^\infty \binom{n+k}{n} x^k = \frac1{(1-x)^{n+1}}.$ Plug in $x = 1/n$ to get $\lim\limits_{n\to\infty} \sum_{k=0}^\infty \frac{\binom{n+k}{n}}{n^k} = \lim_{n\to\infty} \frac1{(1-1/n)^{n+1}},$ and standard techniques show that this limit is $\fbox{e}.$

Without any rigorous justification (see note below) I applied the "limit of a sum is the sum of limits" idea and interchanged the limit and the sum. Then looking at a single one of those limits

$\begin{aligned} \displaystyle \lim_{n \to \infty} \dfrac{ \binom{n+k}{n}}{n^k} &= \displaystyle \lim_{n \to \infty} \dfrac{ \left( \dfrac{(n+k)(n+k-1)(n+k-2) \ldots (n+2)(n+1)}{k!} \right)}{n^k} \\ \\ &= \displaystyle \lim_{n \to \infty} \dfrac{n^k + \binom{k}{2} n^{k-1} + \ldots}{k! \ n^k} \\ \\ &= \dfrac{1}{k!} \end{aligned}$

Then we get $\displaystyle \lim_{n \to \infty} \left( \displaystyle \sum_{k=0}^{\infty} \dfrac{ \binom{n+k}{n}}{n^k} \right) = \displaystyle \sum_{k=0}^{\infty} \left( \displaystyle \lim_{n \to \infty} \dfrac{ \binom{n+k}{n}}{n^k} \right) = \displaystyle \sum_{k=0}^{\infty} \dfrac{1}{k!} = \boxed{e}$

I believe the "limit of a sum is the sum of limits" idea generally only applies if the number of terms in the sum is finite and independent of $n$ , the former of which is not true here. I think the solution still works, but don't know how to justify that first step. Perhaps uniform convergence or something like that, but I have very little understanding of such ideas and whether any might apply here. Can anybody help with that part?