An Eulerian numbers identity proof by induction

Question

The Eulerian numbers $E(n, k)$ satisfy the following recurrence relation:

\begin{equation} \left\langle \begin{matrix} n\\k \end{matrix} \right\rangle = (k+1) \left\langle \begin{matrix} n-1\\k \end{matrix} \right\rangle + (n-k) \left\langle \begin{matrix} n-1 \\k-1 \end{matrix} \right\rangle. \end{equation}

The binomial coefficients satisfy the recurrence:

\begin{align} \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. \end{align}

Recently, I came up with the identity \begin{equation} \left\langle \begin{matrix} n\\k \end{matrix} \right\rangle = (k+1)^{n} - k^{n} \binom{n+1}{n} + (k-1)^{n} \binom{n+1}{n-1} - \cdots + (-1)^{k} \binom{n+1}{n+1-k}, \end{equation} but I am not sure how to prove it with indutcion.

What would be the proof by induction of the identity?

Answer 1

$$\left\langle\begin{matrix}n\\k\end{matrix}\right\rangle=\sum_{j=0}^{k}(-1)^j\binom{n+1}{n+1-j}(k+1-j)^n\tag1$$

Let $L(n,k)$ be the LHS of $(1)$, and $R(n,k)$ be the RHS of $(1)$.

Let us prove $(1)$ by induction on $n$.

We have $L(0,0)=R(0,0)$ and $L(1,0)=R(1,0)$.

Suppose that for $0\le k\lt n$, $(1)$ holds.

Then, we have $$\begin{align}&L(n+1,k)\\\\&=\left\langle\begin{matrix}n+1\\k\end{matrix}\right\rangle \\\\&=(k+1)\left\langle\begin{matrix}n\\k\end{matrix}\right\rangle+(n+1-k)\left\langle\begin{matrix}n\\k-1\end{matrix}\right\rangle \\\\&=(k+1)\sum_{j=0}^{k}(-1)^j\binom{n+1}{n+1-j}(k+1-j)^n+(n+1-k)\sum_{j=0}^{k-1}(-1)^j\binom{n+1}{n+1-j}(k-j)^n \\\\&=(k+1)\bigg((k+1)^n+\sum_{j=\color{red}1}^{k}(-1)^j\binom{n+1}{n+1-j}(k+1-j)^n\bigg)+(n+1-k)\sum_{j=0}^{k-1}(-1)^j\binom{n+1}{n+1-j}(k-j)^n \\\\&=(k+1)^{n+1}+(k+1)\sum_{j=1}^{k}(-1)^j\binom{n+1}{n+1-j}(k+1-j)^n+(n+1-k)\sum_{j=0}^{k-1}(-1)^j\binom{n+1}{n+1-j}(k-j)^n \\\\&=(k+1)^{n+1}+(k+1)\sum_{j=0}^{k-1}(-1)^{j+1}\binom{n+1}{n-j}(k-j)^n+(n+1-k)\sum_{j=0}^{k-1}(-1)^j\binom{n+1}{n+1-j}(k-j)^n \\\\&=(k+1)^{n+1}+\sum_{j=0}^{k-1}(-1)^{j}(k-j)^n\bigg(-(k+1)\binom{n+1}{n-j}+(n+1-k)\binom{n+1}{n+1-j}\bigg) \\\\&=(k+1)^{n+1}+\sum_{j=0}^{k-1}(-1)^{j}(k-j)^n\bigg(-(k+1)\times\frac{(n+1)!}{(n-j)!(j+1)!}+(n+1-k)\times\frac{(n+1)!}{(n+1-j)!j!}\bigg) \\\\&=(k+1)^{n+1}+\sum_{j=0}^{k-1}(-1)^{j}(k-j)^n\frac{(n+1)!}{(n+1-j)!(j+1)!}\bigg(-(k+1)(n+1-j)+(n+1-k)(j+1)\bigg) \\\\&=(k+1)^{n+1}+\sum_{j=0}^{k-1}(-1)^{j}(k-j)^n\frac{(n+1)!}{(n+1-j)!(j+1)!}\times (n+2)(j-k) \\\\&=(k+1)^{n+1}+\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{n+1}\binom{n+2}{j+1} \\\\&=(k+1)^{n+1}+\sum_{j=0}^{\color{red}k}(-1)^{j+1}(k-j)^{n+1}\binom{n+2}{j+1} \\\\&=(k+1)^{n+1}+\sum_{j=1}^{k+1}(-1)^{j}(k-j+1)^{n+1}\binom{n+2}{j} \\\\&=\sum_{j=\color{red}0}^{k+1}(-1)^{j}(k-j+1)^{n+1}\binom{n+2}{j} \\\\&=\sum_{j=0}^{k}(-1)^{j}(k-j+1)^{n+1}\binom{n+2}{j} \\\\&=\sum_{j=0}^{k}(-1)^{j}(k-j+1)^{n+1}\binom{n+2}{n+2-j} \\\\&=R(n+1,k).\ \blacksquare\end{align}$$