Calculating a sum in the coefficient of some generating function

Question

I tried to calculate the coefficient of a generating function \begin{align*} \frac{1}{n} [u^{n-1}] e^{un} \frac{1}{(1-u)^2} \end{align*} and got to \begin{align*} \frac{1}{n} [u^{n-1}] e^{un} \frac{1}{(1-u)^2} &= \frac{1}{n} [u^{n-1}] \Bigg(\sum_{k\geq 0}\frac{1}{k!} (un)^k\Bigg) \Bigg(\sum_{k\geq 0} \underbrace{\binom{n+2-1}{2-1}}_{=~n+1} u^n\Bigg) \\ &= \frac{1}{n} [u^{n-1}] \sum_{k\geq 0} \sum_{l=0}^k \frac{n^l}{l!} (k-l+1) u^{k} = \frac{1}{n} \sum_{l=0}^{n-1} \frac{n^l}{l!} (n-l). \end{align*} According to wolfram alpha, this sum equals \begin{align*} \frac{1}{n} \sum_{l=0}^{n-1} \frac{n^l}{l!} (n-l) = \frac{n^n}{n!}. \end{align*} Is there a quick and simple way to show this or could I have done a different calculation in the beginning to arrive at this solution?

Answer 1

One could also work with geometric series instead of using the general binomial theorem: \begin{align*} [z^n]g(f(z)) &= \frac{1}{n}[u^{n-1}]\phi(u)^n g'(u) \\ &= \frac{1}{n}[u^{n-1}]e^{nu} \left( \frac{1}{1-u}\right)' \\ &= \frac{1}{n}[u^{n-1}]e^{nu} \left( \sum_{k \geq 0}u^k\right)' \\ &= \frac{1}{n}[u^{n-1}]\sum_{k \geq 0}n^k\frac{u^k}{k!} \sum_{k \geq 1}ku^{k-1} \\ &= \frac{1}{n}[u^{n-1}]\sum_{k \geq 0}n^k\frac{u^k}{k!} \sum_{k \geq 0}(k+1)u^{k} \\ &= \frac{1}{n}[u^{n-1}]\sum_{k \geq 0}\sum^{k}_{l = 0}{(k-l+1) \frac{n^l}{l!}} u^k \\ &= \frac{1}{n} \sum^{n-1}_{l = 0}{(n-1-l+1) \frac{n^l}{l!}} \\ &= \frac{1}{n} \sum^{n-1}_{l = 0}{(n-l) \frac{n^l}{l!}} \\ &= \frac{n^n}{n!}. \end{align*} where at the end we used \begin{align} \frac{1}{n} \sum^{n-1}_{l=0}{\frac{n^l}{l!}(n-l)} &= \frac{1}{n}\sum^{n-1}_{l=1}\frac{n^l}{l!}(n-l) + 1 \\ &= \sum^{n-1}_{l=1}\frac{n^l}{l!} - \sum^{n-1}_{l=1}{\frac{n^{l-1}}{(l-1)!}} + 1\\ &= \sum^{n-1}_{l=1}\frac{n^l}{l!} - \sum^{n-2}_{l=0}{\frac{n^{l}}{l!}} + 1 \\ &= \frac{n^{n-1}}{(n-1)!} \\ &= \frac{n^{n-1}n }{n(n-1)!} \\ &= \frac{n^n}{n!} \end{align}