# A Proof of Tepper’s identity

Tepper’s identity. For any real number $a$ we have the following identity $\sum_{k = 0}^n( - 1)^k\binom{n}{k}(a - k)^n = n!\, \forall n\in\mathbb{N}$.

Proof. WLOG, assume that $a = 0$ and $n\geq 3$, then we need to prove $\sum_{k = 0}^n( - 1)^{n + k}\binom{n}{k}( k)^n = n!\, (*)$. By Lagrange’s Interpolation Formula, we have $x^n - (x - 1)(x - 2)\cdots (x - n) \equiv \sum_{k = 1}^nk^n\cdot\prod_{i\not = k}\dfrac{x - i}{k - i}\, (**)$. Now, in $(**)$, setting $x = 0$ we have $( - 1)^{n + 1}\cdot n! = \sum_{k = 1}^nk^n\cdot \dfrac{1}{k}\cdot \dfrac{( - 1)^{n - 1}\cdot n!}{(k - 1)!\cdot (n - k)!\cdot ( - 1)^{n - k}}$, from this and $\binom{n}{k} = \dfrac{n!}{k!\cdot (n - k)!}$ we have $(*)$ is true. $\Box$

## One thought on “A Proof of Tepper’s identity”

1. masteranza says:

It’s great to find this kind of blog in a “field of weeds” 🙂 I’ve never seen before this identity…
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