How can I find a closed form for this sum $\sum_{k=1}^n{n \choose k}p^{k+1}(1-p)^{n-1}$? Based on some guess and check I think that
$$\sum_{k=1}^n{n \choose k}p^{k+1}(1-p)^{n-1}=p(1-p)^{n-1}((p+1)^n-1)$$
Where $0\leq p \leq 1$
but I'm not sure how to get from one to the other, or if it is truly correct.

By the Binomial Theorem, we have that
$$\begin{align}
\sum_{k=1}^{n}\binom{n}{k}p^{k+1}(1-p)^{n-1}&=p(1-p)^{n-1}\sum_{k=1}^{n}\binom{n}{k}p^{k}\\
&=p(1-p)^{n-1}\sum_{k=1}^{n}\binom{n}{k}p^{k}(1)^{n-k}\\
&=p(1-p)^{n-1}\left[\sum_{k=0}^{n}\binom{n}{k}p^{k}(1)^{n-k}-1\right] \quad \text{ since }\binom{n}{0}p^{0}(1)^{n-0}=1 \\
&= p(1-p)^{n-1}[(1+p)^{n}-1]\text{.}
\end{align}$$
