Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let n=0. Then $\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$. Thus for $n=0$ our equation is satisfied.
Inductive Step. Let $n\ge0$. Assume $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$. Now observe that
$$\begin{align*}\sum_{k=0}^{n+1}k{m+k \choose m}&=\\
\sum_{k=0}^{n}k{m+k \choose m}+(n+1){m+n+1\choose m}&=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}+(n+1){m+n+1\choose m} \end{align*}$$
This is where I'm getting stuck... using Pascal's Identity to combine some of these terms seems ideal. However, the factors of $n$ and $n+1$ are making a clever manipulation difficult for me.

$$\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$$
Inductive Step. 1:  $n\longrightarrow0$
$\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$.
the equation is satisfied.
Inductive Step. 2:  $n\longrightarrow n$
$$\sum_{k=0}^nk{m+k \choose m}=$$
By Pascal's identity: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} \Rightarrow \binom{n}{k} - \binom{n-1}{k}= \binom{n-1}{k-1}  \Rightarrow  \binom{n+1}{k+1} - \binom{n-1+1}{k+1}= \binom{n-1+1}{k-1+1}= $
$\binom{n+1}{k+1} - \binom{n}{k+1}= \binom{n}{k}$
$$\sum_{k=0}^nk\left[\binom{m+k+1}{m+1}-\binom{m+k}{m+1}\right]=$$
$$\sum_{k=0}^nk\binom{m+k+1}{m+1}-\sum_{k=0}^nk\binom{m+k}{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n-1}(k+1)\binom{m+(k+1)}{m+1} + 0\binom{m+ 0 }{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n-1}(k+1)\binom{m+k+1}{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}(k-(k+1))\binom{m+k+1}{m+1}=$$
$$n\binom{m+n+1}{m+1}+\sum_{k=0}^{n-1}-1\binom{m+k+1}{m+1}=$$
By the Hockey-Stick identity:
$\sum_{k=r}^n\binom{k}{r}=\binom{n+1}{r+1}$
$$n\binom{m+n+1}{m+1}-\binom{m+(n-1)+1 +1}{m+2}=$$
$$n\binom{m+n+1}{m+1}-\binom{m+n+1}{m+2}$$
The equation is satisfied.
Inductive Step. 3:  $n\longrightarrow (n+1)$
$$\sum_{k=0}^{(n+1)}k{m+k \choose m}=$$
$$\sum_{k=0}^{(n+1)}k\left[\binom{m+k+1}{m+1}-\binom{m+k}{m+1}\right]=$$
$$\sum_{k=0}^{(n+1)}k\binom{m+k+1}{m+1}-\sum_{k=0}^{(n+1)}k\binom{m+k}{m+1}=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n}(k+1)\binom{m+(k+1)}{m+1} + 0=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}k\binom{m+k+1}{m+1}-\sum_{k=0}^{n}(k+1)\binom{m+k+1}{m+1}=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}(k-(k+1))\binom{m+k+1}{m+1}=$$
$$(n+1)\binom{m+(n+1)+1}{m+1}+\sum_{k=0}^{n}-1\binom{m+k+1}{m+1}=$$
By the Hockey-Stick identity:
$\sum_{k=r}^n\binom{k}{r}=\binom{n+1}{r+1}$
$$(n+1)\binom{m+(n+1)+1}{m+1}-\binom{m+(n+1) +1}{m+2}$$
The equation is satisfied.
q.e.d
