Prove algebraically
$\sum_{k=m}^{n} k^{\downarrow m }{n \choose k} = n^{\downarrow m } 2^{n-m}$
I have an idea as to how to prove it when m = 1, but am having trouble otherwise.
When m=1, we just have
$\sum_{k=1}^{n} k^{\downarrow 1 }{n \choose k} = n^{\downarrow 1 } 2^{n-1} = n2^{n-1} $
I would appreciate any insight you guys could give would be appreciated. I do understand the identity combinatorially but am having issues with the algebra.
Longtime lurker, 1st time poster, so please forgive if I am not following proper protocol.