real analysis - Sum of infinite series $sum_{n=0}^{infty}(-1)^nfrac{n-1}{n!}t^n$

I have this problem, finding infinite sum of this series

$$\sum_{n=0}^{\infty}(-1)^n\frac{n-1}{n!}t^n$$

It should be done using derivatives and integrals, like for example:

$$\sum_{n=1}^{\infty}\frac{t^n}{n}=\sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}=\sum_{n=0}^{\infty}\int_{0}^{t}s^nds=\int_{0}^{t}\sum_{n=0}^{\infty}s^nds=\int_{0}^{t}\frac{1}{1-s}ds=-ln(1-t)$$

I have some ideas on how the solution should be found, but this $(-1)^n$ is keeping me confused, I don't know what should I do with it.

Any help would be very appreciated. Thanks!

Answer

$$\ldots =\sum_{n=0}^\infty \frac{n}{n!} (-t)^n - \sum_{n=0}^\infty \frac{1}{n!}(-t)^n=\sum_{n=1}^\infty \frac{1}{(n-1)!} (-t)^{n} -e^{-t}=(-t) e^{-t} - e^{-t}.$$