How to prove $\sum_{k=1}^{n}kx^{k} \binom{n}{k}= nx(1+x)^{n-1}$? How to prove $\sum_{k=1}^{n}kx^{k}\binom{n}{k}= nx(1+x)^{n-1}$?
I can see the LHS has some structure to it and if we don't consider the $k$ term then we have the binomial formula for $(1+x)^{n}$ but the $k$ makes things complicated for me. I tried to shift the $k$ values around but I am not exactly sure how I can use that though? And I feel like we can use the fact that the RHS has $(1+x)^{n-1}$. But how?
Any hints/suggestions?

\begin{align}
\sum\limits_{k = 1}^n {kx^k \binom{n}{k}} &= \sum\limits_{k = 1}^n {kx^k \frac{{n!}}{{(n - k)!k!}}}\\ 
&= nx\sum\limits_{k = 1}^n {x^{k - 1} \frac{{(n - 1)!}}{{(n - k)!(k - 1)!}}} \\ 
&= nx\sum\limits_{k = 1}^n {x^{k - 1} \frac{{(n - 1)!}}{{(n - 1 - (k - 1))!(k - 1)!}}}\\
&= nx\sum\limits_{k = 0}^{n-1} {x^{k} \frac{{(n - 1)!}}{{(n - 1 - k)!k!}}} \\ &= nx\sum\limits_{k = 0}^{n-1} {x^{k} \binom{n-1}{k}}\\ 
&= nx(1 + x)^{n - 1} .
\end{align}
