Value of convergent infinite factorial series

I've tried searching for this both with "infinite factorial series" and on approach0.xyz to no avail.

Wolfram|Alpha gives:
$$\sum_{i=0}^{\infty}\frac{i!}{(i+4)!}=\frac{1}{18}$$
I was wondering where this came from. It's the infinite limit of the partial sum formula given by Wolfram|Alpha, as expected, but that also raises the question of where the partial sum formula comes from.

Any help would be appreciated.

Answer

Noting
$$ \frac{1}{(x+1) (x+2) (x+3) (x+4)}=-\frac12\bigg(\frac{1}{x+2}-\frac{1}{x+3}\bigg)+\frac{1}{6}\bigg(\frac1{x+1}-\frac{1}{x+4}\bigg)$$

one has
\begin{eqnarray}
\sum_{i=0}^{\infty}\frac{i!}{(i+4)!}&=&\sum_{i=0}^{\infty}\frac{1}{(i+1)(i+2)(i+3)(i+4)}\\
&=&-\frac12\sum_{i=0}^{\infty}\bigg(\frac{1}{i+2}-\frac{1}{i+3}\bigg)+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{x+1}-\frac{1}{x+4}\bigg)\\
&=&-\frac12\cdot\frac12+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{x+1}-\frac{1}{x+2}\bigg)+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{x+2}-\frac{1}{x+3}\bigg)+\frac16\sum_{i=0}^{\infty}\bigg(\frac1{x+3}-\frac{1}{x+4}\bigg)\\
&=&-\frac14+\frac16(1+\frac12+\frac13)\\
&=&\frac1{18}.
\end{eqnarray}