summation - Calculating the sum $sum_{n=1}^infty frac{n}{(n-1)!}x^n$?

So I've got the sum $$\sum_{n=1}^\infty \frac{n}{(n-1)!}x^n$$

To show that it converges for all real numbers, I used the ratio test. And found the convergence radius to be $$R = \frac{1}{L}, \qquad R = \infty$$
The next task is to calculate the sum, and I feel sort of lost.. I think I want the sum too look like a geometric series. Or substitute it with something else.

Answer

Recall that
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}.$$
First way. Note that
$$xe^x=x(e^x)'=x\sum_{n=0}^\infty n\frac{x^{n-1}}{n!}=\sum_{n=1}^\infty \frac{x^{n}}{(n-1)!}.$$
Try to differentiate again and compare the result with your series.

Second way. we have that

$$\sum_{n=1}^\infty \frac{n}{(n-1)!}x^n=\sum_{m=0}^\infty \frac{m+1}{m!}x^{m+1}=x^2\sum_{m=1}^\infty \frac{m}{m!}x^{m-1}+x\sum_{m=0}^\infty \frac{x^{m}}{m!}.$$
What then?