Proving $\binom{n+1+m}{n+1}=\sum_{k=0}^m(k+1)\binom{n+m-k}{n}$ 
Use any method to prove that 
  $$\binom{n+1+m}{n+1}=\sum_{k=0}^m(k+1)\binom{n+m-k}{n}$$

My Try:
Base case: Let $m=1$
LHS$$\binom{n+1+m}{n+1}=\binom{n+2}{n+1}=(n+2)$$
RHS$$\sum_{k=0}^m(k+1)\binom{n+m-k}{n}=\binom{n+1-0}{n}+(1+1)\binom{n+1-1}{n}$$
$$=\frac{(n+1)!}{n!}+2$$
$$=(n+3)$$
If $m=2$
LHS$$\binom{n+3}{n+1}=\frac{n^2+5n+6}{2!}$$
RHS$$=\binom{n+2}{n}+2\binom{n+1}{n}+3\binom{n}{n}$$
$$=\frac{n^2+3n+2+4n+4+6}{2}=\frac{n^2+7n+12}{2}$$
Clearly $LHS\ne RHS$
If LHS and RHS are not equal then how to prove this proof? Can anyone explain how to prove this.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\require{cancel}\bcancel{\cancel{n + 1 + m \choose n + 1}} =
\sum_{k = 0}^{m}\pars{k + 1}{n + m - k \choose n}:\ {\LARGE ?}}$.
  The right answer is
$\bbx{\ds{n + m + 2 \choose n + 2}}$

$$
\bbx{\mbox{Note that}\ {n + m - k \choose n} = 0\
\mbox{when}\ k > m}
$$

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 0}^{m}\pars{k + 1}
{n + m - k \choose n}} =
\sum_{k = 0}^{\infty}\pars{k + 1}{n + m - k \choose m - k} 
\\[5mm] = &\
\sum_{k = 0}^{\infty}\pars{k + 1}
\bracks{{-n - 1 \choose m - k}\pars{-1}^{m - k}}
\\[5mm] = &\
\pars{-1}^{m}\sum_{k = 0}^{\infty}\pars{k + 1}\pars{-1}^{k}
\bracks{z^{m - k}}\pars{1 + z}^{-n - 1}
\\[5mm] = &\
\pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 1}
\sum_{k = 0}^{\infty}\pars{k + 1}\pars{-z}^{k}
\\[5mm] = &\
\pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 1}\,
\pars{-\,\partiald{}{z}\sum_{k = 0}^{\infty}\pars{-z}^{k + 1}}
\\[5mm] = &\
\pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 1}\,
\partiald{}{z}\pars{z \over 1 + z} =
\pars{-1}^{m}\bracks{z^{m}}\pars{1 + z}^{-n - 3}
\\[5mm] = &\
\pars{-1}^{m}{-n - 3 \choose m} =
\pars{-1}^{m}\bracks{{n + 3 + m - 1\choose m}\pars{-1}^{m}}
\\[5mm] = &\
\bbx{n + m + 2 \choose n + 2}
\end{align}
A: The RHS can be written as $$\sum_{i+j=n+m+1}\binom{i}1\binom{j}n$$where $\binom{r}{s}:=0$ if $s\notin\{0,\dots,r\}$.
This equals: $$\binom{n+2+m}{n+2}$$
See here for a proof of that. So RHS does not equal LHS.
