partial sum involving factorials Here is an interesting series I ran across.  
It is a binomial-type identity.  
$\displaystyle \sum_{k=0}^{n}\frac{(2n-k)!\cdot 2^{k}}{(n-k)!}=4^{n}\cdot n!$
I tried all sorts of playing around, but could not get it to work out.
This works out the same as $\displaystyle 2^{n}\prod_{k=1}^{n}2k=2^{n}\cdot 2^{n}\cdot n!=4^{n}\cdot n!$
I tried equating these somehow, but I could not get it. I even wrote out the series.
There were cancellations, but it did not look like the product of the even numbers.
$\displaystyle \frac{(2n)!}{n!}+\frac{(2n-1)!\cdot 2}{(n-1)!}+\frac{(2n-2)!2^{2}}{(n-2)!}+\cdot\cdot\cdot +n!\cdot 2^{n}=4^{n}\cdot n!$.
How can the closed form be derived from this?.  I bet I am just being thick. I see the last term is nearly the result except for being multiplied by $2^{n}$.  I see if the factorials are written out, $2n(2n-1)(2n-2)(2n-3)\dots$ for example,  then 2's factor out of $2n, \;\ 2n-2$ (even terms) in the numerator. 
There is even a general form I ran through Maple. It actually gave a closed from for it as well, but I would have no idea how to derive it. 
$\displaystyle \sum_{k=0}^{n}\frac{(2n-k)!\cdot 2^{k}\cdot (k+m)!}{(n-k)!\cdot k!}$.
In the above case, m=0.  But, apparently there is a closed form for $m\in \mathbb{N}$ as well.
Maple gave the solution in terms of Gamma:  $\displaystyle \frac{\Gamma(1+m)4^{n}\Gamma(n+1+\frac{m}{2})}{\Gamma(1+\frac{m}{2})}$
Would anyone have an idea how to proceed with this?.  Perhaps writing it in terms of Gamma and using some identities?. Thanks very much. 

This identity can be re-written as
$$\sum_{k=0}^n {2n-k \choose n-k} 2^k = 4^n.$$
Start from
$${2n-k \choose n-k} =
\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{2n-k}}{z^{n-k+1}} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} 
\sum_{k=0}^n \frac{(1+z)^{2n-k}}{z^{n-k+1}} 2^k \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} 
\frac{(1+z)^{2n}}{z^{n+1}}
\sum_{k=0}^n \frac{(2z)^{k}}{(1+z)^k} \; dz.$$
We can extend the sum to infinity because when $n-k+1 \le 0$ or $k \ge n+1$ the integrand of the defining integral of the binomial coefficient is an entire function and the integral is zero. This yields
$$\frac{1}{2\pi i} \int_{|z|=\epsilon} 
\frac{(1+z)^{2n}}{z^{n+1}}
\sum_{k=0}^\infty \frac{(2z)^{k}}{(1+z)^k} \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} 
\frac{(1+z)^{2n}}{z^{n+1}} \frac{1}{1-2z/(1+z)} \; dz
\\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} 
\frac{(1+z)^{2n+1}}{z^{n+1}} \frac{1}{1-z} \; dz.$$
Thus the value of the integral is given by
$$[z^n] \frac{1}{1-z} (1+z)^{2n+1}
= \sum_{q=0}^n {2n+1\choose q} = \frac{1}{2} 2^{2n+1} = 4^n.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
