Prove that: $\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$ 
Prove that:
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$

My work so far:
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
$$\frac{(n+0)!}{n!0!}+\frac{(n+1)!}{n!1!}+...+\frac{(n+n)!}{n!n!}=\frac{(2n+1)!}{n!(n+1)!}$$
$$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+...+\binom{2n}{n}=\binom{2n+1}{n}$$
How to prove the last equality?
 A: Combinatorial proof of
$$\frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}+...+\frac{(n+n)!}{n!}=\frac{(2n+1)!}{(n+1)!}$$
that is, after dividing by $n!$,
$$\binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{2n}{n}=\binom{2n+1}{n+1}.$$
The RHS counts the number of $\{0,1\}$-strings  of length $2n+1$ with $(n+1)$ $1$s. The LHS enumerates the same set according to the position $k$ of the last $1$ on the right: for $k=n+1,n+2,\dots, 2n+1$ they are
$$\binom{k-1}{n}.$$
A: Try to use the following identity:
$$
\binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}.
$$
A: First, note ${n\choose0}={n+1\choose0}$, then repeatedly use ${k\choose r}+{k\choose r+1}={k+1\choose r+1}$
A: $$\sum_{r=0}^n\binom{n+r}r$$  is the coefficient of $x^n$  in $$\sum_{r=0}^n(1+x)^{n+r}$$ which is a Finite Geometric Series and is  $$=(1+x)^n\cdot\dfrac{(1+x)^{n+1}-1}{1+x-1}=\dfrac{(1+x)^{2n+1}-(1+x)^n}x$$
Now the coefficient of $x^{n+1}$ in $$(1+x)^{2n+1}-(1+x)^n$$
$$=\binom{2n+1}{n+1}-0$$
A: You can calculate directly:
$$\begin{align} & \frac{(n+0)!}{0!}+\frac{(n+1)!}{1!}=\frac{n!(1+n+1)}{1!}; \\
& \frac{n!(n+2)}{1!}+\frac{(n+2)!}{2!}=\frac{n!(n+2)(2+n+1)}{2!}; \\
& \cdots \\
& \frac{n!(n+2)(n+3)\cdots ((n-1)+n+1)}{(n-1)!}+\frac{(n+n)!}{n!}= \\ 
& \frac{n!(n+2)(n+3)\cdots (n+n+1)}{n!}=\frac{(2n+1)!}{(n+1)!}.
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 0}^{n}{\pars{n + k}! \over k!} & =
n!\sum_{k = 0}^{n}{n + k \choose k} =
n!\sum_{k = 0}^{n}{-n - 1 \choose k}\pars{-1}^{k} =
n!\sum_{k = 0}^{n}\pars{-1}^{k}{-n - 1 \choose -n - 1 - k}
\\[5mm] & =
n!\sum_{k = 0}^{n}\pars{-1}^{k}\bracks{z^{-n - 1 - k}}
\pars{1 + z}^{-n - 1} =
n!\bracks{z^{-n - 1}}{1 \over \pars{1 + z}^{n + 1}}
\sum_{k = 0}^{n}\pars{-z}^{k}
\\[5mm] & =
n!\bracks{z^{-n - 1}}{1 \over \pars{1 + z}^{n + 1}}\,
{\pars{-z}^{n + 1} - 1 \over -z - 1} =
-n!\bracks{z^{-n - 1}}
{1 - \pars{-1}^{n}z^{n + 1} \over \pars{1 + z}^{n + 2}}
\\[5mm] & =
-n!\bracks{z^{n + 1}}z
{z^{n + 1} - \pars{-1}^{n} \over \pars{1 + z}^{n + 2}} =
{\pars{-1}^{n} \over n!}\bracks{z^{n}}\pars{1 + z}^{-n - 2} =
{\pars{-1}^{n} \over n!}{-n - 2 \choose n}
\\[5mm] & =
\pars{-1}^{n}\,n!{2n + 1 \choose n}\pars{-1}^{n} = \bbx{\pars{2n + 1}! \over \pars{n + 1}!}
\end{align}
