TITLE: Prove $\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n = (n+2)^n$
QUESTION [10 upvotes]: I found through simulations that 

$$\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n  = (n+2)^n$$

Is there any proof of this?  I've tried to solve it by:

Induction, but it gets too messy.
Binomial theorem, but I got nowhere.

Any help with this is highly appreciated.

REPLY [4 votes]: This is a less clever algebraic proof by induction. We actually prove something stronger:

For any $\lambda>0$, we have
  $$\sum_i(-1)^i\binom{n+1}{i}(i+\lambda)^n=0$$
  where $i$ runs through all integers and define $\binom{n}{i}=0$ if $i<0$ or $i>n$.

Proof. This is straight-forward for $n=0$. Suppose for $n=k$ the statement is true. Then
\begin{align}
\sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=\sum_i(-1)^i\binom{k+2}i\binom i1(i+\lambda)^k+\lambda\sum_i(-1)^i\binom{k+2}{i}(i+\lambda)^k
\end{align}
Note that $\binom{k+2}i\binom i1=\binom{k+2}1\binom{k+1}{i-1}$ and $\binom{k+2}i=\binom{k+1}i+\binom{k+1}{i-1}$. By induction hypothesis we obtain $\sum(-1)^i\binom{k+1}i(i+\lambda)^k=0$, thus
\begin{align}
\sum_i(-1)^i\binom{k+2}i(i+\lambda)^{k+1}&=(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i-1}(i+\lambda)^k\\
&=-(k+2+\lambda)\sum_i(-1)^i\binom{k+1}{i}(i+\lambda+1)^k\\
&=0
\end{align}
where the last equality follows from induction hypothesis again.