Prove that $ \sum_{m=0}^{{\rm min}\{n,l\}} \binom{l}{m} \frac{n!}{(n-m)!} x^{l-m} = \sum_{m=0}^{l} \binom{l}{m} (n-l+m+1)_{l-m} x^{m}$ How to show that for any non-negative integers $n$ and $l$ 
two sums below are equal:
$$
\sum_{m=0}^{{\rm min}\{n,l\}} \binom{l}{m} \frac{n!}{(n-m)!} x^{l-m}
= \sum_{m=0}^{l} \binom{l}{m} (n-l+m+1)_{l-m}  x^{m}, $$
where $(a)_s= a(a+1)(a+2)⋯(a+n−1)$ is the Pochhammer symbol.
 A: Let $k_0=\operatorname{min}\{n,l\}$. Then observe that the co-efficient of $x^k \neq 0$ on the LHS if and only if $l-k_0\leq k\leq l$.
So, we first assume $l-k_0\leq k \leq l.$ Now, observing $n-(l-k)\geq 0$, we have,
$$\begin{align}
\ {\rm coefficient \space of} \space x^k \space {\rm on \space the \space RHS} & = {l \choose k}(n-l+k+1)_{l-k}\\
& =\ {l \choose k}\prod_{i=1}^{l-k}(n-(l-k)+i)\\
&=\ {l \choose k}\left(\frac{(n-(l-k))!\left(\prod_{i=1}^{l-k}(n-(l-k)+i)\right)}{(n-(l-k))!}\right)\\
&=\ {l \choose k}\frac{n!}{(n-(l-k))!}\\
&=\ {l \choose l-k}\frac{n!}{(n-(l-k))!}\\
&=\ {\rm coefficient \space of} \space x^k \space {\rm on \space the \space LHS}.
\end{align}$$
Now we assume $k<l-k_0$. If $k_0=l$, then $k<0$ and thus the coefficient of $x^k=0$ both on the LHS and the RHS. So we assume $k_0=n$.
In that case observe $\require{cancel}n-l+k+1<n-\cancel{l}+\cancel{l}-k_0+1=n-n+1=1$. Thus, $n-l+k+1 \leq 0$ and as $n\geq 0$,
$$0 \in \{n-l+k+1+i\mid 0 \leq i \leq l-k-1\}$$
and hence $(n-l+k+1)_{l-k}=\prod_{i=0}^{l-k-1}(n-l+k+1+i)=0.$
Thus, coefficient of $x^k$ on the RHS $={l \choose k}(n-l+k+1)_{l-k}=0=$ coefficient of $x^k$ on the LHS. 
A: Equating similar powers of $x$:
$$l-m = k$$
also $$ \binom {l}{m} = \binom {l}{(l-m)} = \binom {l}{k}$$
now,
$$(a)_n = a(a+1)...(a+n-1) = \frac{(a+n-1)!}{(a-1)!}$$
So, $$(n-l+k+1)_{(l-k)} = (n-m+1)_m = \frac{n!}{(n-m+1-1)!} = \frac{n!}{(n-m)!}$$
Thus, for all powers the 2 terms on LHS and RHS are equal.
All that is left now is to ensure that the number of terms is same.
case 1) Let $min(n, l)=l$ then number of terms are obviously same on LHS and RHS.
case 2) if $min(n, l) = n$ then LHS terms are from $x^l$ to $x^{l-n}$. On RHS, $((n-l)+(m+1))_{(l-m)}$ is valid only if $(n-l+m) \ge 0$ or $m \ge (l-n)$. So, the RHS summation runs from $(l-n)$ to $l$ or the terms go from $x^{(l-n)}$ to $x^l$
