I have the following finite summation: $$\sum_{i=0}^{n}(-1)^{i+n} i^n {n \choose i}$$ that I've proven $\forall n \in \mathbb{N}$ to be equal to $n!$. However, I've noticed that if I introduce a constant $a$ into this summation as follows:$$\sum_{i=0}^{n} (-1)^{i+n} (i+a)^n {n \choose i}$$ that the $a$'s all cancel out, and we are left again with $n!$ (I know this from Python and WolframAlpha, not from mathematical proof). Is there any way to prove that this second summation is equivalent to the first one?
That is, show that the second summation is equal to $n!$ by proving the first summation is equal to the second.
I'm looking to show that the introduction of the $(i+a)$ does not affect the summation, and I want to show that the $+a$ will cancel out using the rules for manipulating summations.
I would greatly appreciate any help anyone could provide.