Evaluate $\sum_{n=1}^{\infty} \frac{n^{5}}{n!}$.
Solution: We start by noticing that $\sum_{n=1}^{\infty} \frac{n}{n!}=\sum_{n=1}^{\infty} \frac{1}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1}{n!}=e$. Next we see that $\sum_{n=1}^{\infty} \frac{n^{2}}{n!}=\sum_{n=1}^{\infty} \frac{n}{(n-1)!}=\sum_{n=0}^{\infty} \frac{1+n}{n!}=\sum_{n=0}^{\infty} \frac{1}{n!}+\sum_{n=0}^{\infty} \frac{n}{n!}=e+e=2 e$. Let $f(k)=\sum_{n=1}^{\infty} \frac{n^{k}}{n!}$, then $\sum_{n=1}^{\infty} \frac{n^{k}}{n!}=\sum_{n=1}^{\infty} \frac{n^{k-1}}{(n-1)!}=\sum_{n=0}^{\infty} \frac{(1+n)^{k-1}}{n!}$, so by the binomial theorem $f(k)=\sum_{j=0}^{k-1}\binom{k-1}{j} \cdot f(j)$. Armed with this formula, we can easily compute $f(3)=f(0)+2 f(1)+f(2)=e+2 e+2 e=5 e$, $f(4)=1 \cdot e+3 \cdot e+3 \cdot 2 e+1 \cdot 5 e=15 e$, and $f(5)=1 \cdot e+4 \cdot e+6 \cdot 2 e+4 \cdot 5 e+1 \cdot 15 e=52 e$.