It's a Euler's sum

Let $S$ be the sum in question. Then $\begin{aligned} S&=\sum_{n=2}^\infty \frac{n^2}{n! (n-1)n} -\sum_{n=2}^\infty \frac{n-1}{n! (n-1)n} \\ &=\sum_{n=2}^\infty \frac{1}{(n-1)! (n-1)} -\sum_{n=2}^\infty \frac{1}{n! n} \\ &=\sum_{n=1}^\infty \frac{1}{n! n} -\sum_{n=2}^\infty \frac{1}{n! n} \\ &=\frac{1}{1! \cdot 1} \\ &=\boxed1 \end{aligned}$

Just substitute $n = 2$ to that fraction.