$$e^x-1=\sum_{n\geq1}\frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=\sum_{n\geq1}\frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=\sum_{n\geq1}\frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=\sum_{n\geq1}\frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$\sum_{n\geq1}\frac{n^2}{n!}=2e$$
Edit
This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $x\frac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=\sum_{n\geq1}\frac{x^n}{n!}$$
Apply $x\frac{d}{dx}$:
$$xe^x=\sum_{n\geq1}\frac{x}{n!}x^n$$
Apply $x\frac{d}{dx}$:
$$x(x+1)e^x=\sum_{n\geq1}\frac{n^2}{n!}x^n$$
The pattern continues:
$$\left(x\frac{d}{dx}\right)^k[e^x-1]=\sum_{n\geq1}\frac{n^k}{n!}x^n$$
A similar thing can be done with integration. Example:
Evaluate $$S=\sum_{n\geq0}\frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$\frac1{1+t^2}=\sum_{n\geq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$\arctan x=\sum_{n\geq0}\frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=\sum_{n\geq0}\frac{(-1)^n}{(2n+2)(2n+1)}=\frac\pi4-\frac12\log2$$