Evaluate $\sum^{n-1}_{k=1} {n\choose k}\frac{kn^{n-k}}{k+1}$ How can we evaluate:

$$\sum^{n-1}_{k=1} {n\choose k}\frac{kn^{n-k}}{k+1}$$

I observe that the expression has a similar format with the following formula:
$$\sum^{n}_{j=0} {n\choose j}\frac{u^j}{j+1}=\frac{1}{n+1}\sum^n_{k=0}(u+1)^k$$ but I have no idea how to process further. A blind evaluation using Wolfram Alpha gives us $\frac{n(n^n-1)}{n+1}$

$\displaystyle \sum\limits^{n-1}_{k=1} {n\choose k}\frac{k x^{n-k}}{k+1}=\sum\limits^{n-1}_{k=1} {n\choose k}x^{n-k}-\frac{1}{n+1}\sum\limits^{n}_{k=2} {n+1\choose k}x^{(n+1)-k}=...$ 
I am sure you can finish - and it's left to set $x:=n$ .
I had used $k=(k+1)-1$ and ${n\choose k}\frac{n+1}{k+1}={n+1\choose k+1}$ .
