Given $n![\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\frac{x^n}{n!} ]^n$, how do I find coefficient of $x^k$ in it ?
How to find coefficient in case of above series having infinite terms i.e. $n! [\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+... ]^n$ ?
So lets consider the infinite case. First note that $x+\frac{x}{2!}+\frac{x}{3!}+...=e^x-1$, and so
\begin{align*} n!\left(x+\frac{x}{2!}+\frac{x}{3!}+...\right)^n=n!(e^x-1)^n. \end{align*}
Using the Binomial Theorem, we have that
\begin{align*} n!(e^x-1)^n & = n!\sum_{i=0}^n \binom{n}{i}e^{x(n-i)}(-1)^{i}\\ & = n!\left(e^{nx}-ne^{x(n-1)}+\binom{n}{2}e^{x(n-2)}-...\pm 1 \right)\\ & = n!\left(\sum_{r=0}^\infty n^r \frac{x^r}{r!}-n\sum_{r=0}^\infty (n-1)^r \frac{x^r}{r!}+\binom{n}{2}\sum_{r=0}^\infty (n-2)^r \frac{x^r}{r!}-...\pm 1\right). \end{align*}
So then it follows that the coefficient for $x^k/k!$ is
\begin{equation} n!\left(n^k-n(n-1)^k+\binom{n}{2}(n-2)^k-...\pm \binom{n}{n-1}\right)=n!\sum_{l=0}^{n-1}(-1)^l\binom{n}{l}(n-l)^k. \end{equation}
So then the coefficient for the $x^k$ would be
\begin{equation} \frac{n!}{k!}\sum_{l=0}^{n-1}(-1)^l\binom{n}{l}(n-l)^k. \end{equation} Note: the finite case follows a similar idea.