A identity which looks like the binomial identity Let $f_0(x)=1$, $f_k(x)=x(x-1)\cdots(x-k+1)$, $\forall k\geq1$. 
For any positive integer $n$, how to prove the following identity:
$$f_n(x+y)=\sum_{k=0}^n\binom{n}{k}f_k(x)f_{n-k}(y)?$$
 A: Let
$$ g_x(z) = \sum_{k\geq 0}x(x-1)\cdot(x-k+1)\frac{z^k}{k!}=\sum_{k\geq 0}\binom{x}{k}z^k=(z+1)^x $$
where the domain of the binomial coefficient $\binom{x}{k}$ has been extended through $\binom{x}{k}=\frac{x!}{k!(x-k)!}=\frac{\Gamma(x+1)}{\Gamma(k+1)\Gamma(x-k+1)}$. It is pretty clear that $g_x(z)\cdot g_y(z)=g_{x+y}(z)$, and the given identity follows from considering the coefficient of $z^n$ in $(z+1)^{x+y}$ and the coefficient of $z^n$ in $g_x(z)\cdot g_y(z)$, regarded as a Cauchy product. As already mentioned in the comments, the statement is just the umbral/q-analog of the binomial theorem.
A: Simple induction will do; I leave the base case for you ...
\begin{eqnarray*}
f_{n+1}(x+y)&=&f_n(x+y) (x+y-n) \\
&=& \sum_{k=0}^{n} \binom{n}{k}f_k(x)f_{n-k}(y) (\color{green}{x-k}+\color{blue}{y-(n-k)}) \\
&=& \sum_{k=0}^{n} \binom{n}{k}f_{k+1}(x)f_{n-k}(y) +\sum_{k=0}^{n} \binom{n}{k}f_k(x)f_{n-k+1}(y) \\
&=& f_{n+1}(x)+\sum_{k=1}^{n} f_{k+1}(x)f_{n-k}(y) \left( \binom{n}{k}+\binom{n}{k-1}\right)+f_{n+1}(y) \\
&=&\sum_{k=0}^{n+1} \binom{n+1}{k}f_{k}(x)f_{n+1-k}(y)  
\end{eqnarray*}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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$\ds{\mrm{f}_{0}\pars{x} \equiv 1\quad\mbox{and}\quad\mrm{f}_{k}\pars{x} =
x\pars{x - 1}\cdots\pars{x - k + 1}\,,\ \forall\ k \in \mathbb{N}_{\geq 1}}$.

  $\ds{\mrm{f}_{n}\pars{x + y} = \sum_{k = 0}^{n}{n \choose k}\mrm{f}_{k}\pars{x}\mrm{f}_{n - k}\pars{y}:\ {\large ?}}$.

Note that $\ds{\mrm{f}_{k}\pars{x} = \pars{x - k + 1}^{\,\large\overline{k}} =
{\Gamma\pars{\bracks{x - k + 1} + k} \over \Gamma\pars{x - k + 1}} =
{x! \over \pars{x - k}!} = k!\,{x \choose k}}$.

\begin{align}
\mrm{f}_{n}\pars{x + y} & =
n!{x + y \choose n} = n!\bracks{z^{n}}\pars{1 + z}^{x + y} =
n!{x + y \choose n} = n!\bracks{z^{n}}\pars{1 + z}^{x}\pars{1 + z}^{y}
\\[5mm] & =
n!\bracks{z^{n}}\bracks{\sum_{i = 0}^{x}{x \choose i}z^{i}}
\bracks{\sum_{j = 0}^{y}{y \choose i}z^{i}} =
n!\sum_{i = 0}^{x}\sum_{j = 0}^{y}\overbrace{{\mrm{f}_{i}\pars{x} \over i!}}
^{\ds{x \choose i}}\ \overbrace{{\mrm{f}_{j}\pars{y} \over j!}}
^{\ds{y \choose j}}\ \bracks{i + j = n}
\\[5mm] & =
\sum_{i = 0}^{x}n!\,{\mrm{f}_{i}\pars{x}\mrm{f}_{n - i}\pars{y} \over i!\ \pars{n - i}!}\bracks{0 \leq n - i \leq y} =
\sum_{i = 0}^{x}{n \choose i}\mrm{f}_{i}\pars{x}\mrm{f}_{n - i}\pars{y}
\bracks{n - y \leq i \leq n}
\\[5mm] & =
\bbx{\bracks{y \geq 0}\sum_{i\ =\ \max\braces{0,n - y}}^{\min\braces{n,x}}
{n \choose i}\mrm{f}_{i}\pars{x}\mrm{f}_{n - i}\pars{y}}
\end{align}
