TITLE: Evaluating $\sum_{k=a}^{15} \frac{k!}{(k-a)!}$ for $0 \leq a \leq 15$
QUESTION [2 upvotes]: I am trying to simplify the sum $\sum_{k=a}^{15} \frac{k!}{(k-a)!}$ for a given integer $a \in [0, 15]$.
Plugged into WolframAlpha, I see that the expression is equivalent to $\frac{16!}{(a+1) \cdot (15 - a)!}$ but I have no clue as to how to get there.
Any help would be appreciated!

REPLY [1 votes]: In general: $$\sum_{k=a}^n\binom{k}{a}=\binom{n+1}{a+1}$$
(For a proof see here)
As @Jean Marie remarked this is the so-called hockey stick identity.
Applying that we find:$$\sum_{k=a}^{15}\frac{k!}{\left(k-a\right)!}=a!\sum_{k=a}^{15}\binom{k}{a}=a!\binom{16}{a+1}=\frac{16!}{\left(a+1\right)\left(15-a\right)!}$$
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