TITLE: Combinatorial and Algebraic Proof of an Identity involving Stirling Numbers of the second kind ${n+1\brace k+1}=\sum_i \binom{n}{i}{i\brace k}$
QUESTION [6 upvotes]: The question is to prove the identity

$$
{n+1\brace k+1}=\sum_i \binom{n}{i}{i\brace k}\tag{1}
$$

via a combinatorial proof and an algebraic proof. The question is from Aigner's A Course in Enumeration. 
The braces indicate Stirling numbers of the second kind. I have managed to prove the identity using the polynomial method (which I will show below), but have not made much progress on the combinatorial proof. 
My attempt: The left hand side represents the number of ways to partition an $n+1$ element set (say $[n+1]=\{1,\dotsc, n+1\}$) into $k+1$ sets. Each of the summands on the right hand side represents choosing $i$ elements where $k \leq i\leq n$ from $[n]$ and then partitioning them into $k$ sets. There is probably some sort of classification of the partition of an $n+1$ element set in to $k+1$ sets but I am not seeing it.
Algebraic Proof:
We expand $(x+1)^n$ in two different ways. First, note that
$$
(x+1)^n=\sum_{i=0}^n\binom{n}{i}x^i=\sum_{i=0}^n\binom{n}{i}\sum_{k=0}^i
{i\brace k} (x)_k=\sum_{k=0}^n(x)_k\left[\sum_{i=k}^n \binom{n}{i}{i\brace k}
\right]\tag{2}
$$ 
by the binomial theorem where $(x)_k$ is the falling factorial of length $k$. For the second way write $(x+1)^n$ as
$$
\sum_{k=0}^n{n\brace k}(x+1)_{k}
=\sum_{k=0}^n{n\brace k}[(x)_k+k(x)_{k-1}]=
\sum_{k=0}^n\left[{n\brace k}+(k+1){n\brace k+1}\right](x)_k\tag{3}
$$
and conclude using the recurrence relation for Stirling numbers.

REPLY [6 votes]: Your algebriac proof is fine. For the Combinatorial proof consider the set that conatins the element $n+1$.
Choose $n-i$ elements from $[n]$ and create a block containing $n+1$ and the $n-i$ chosen elements. Now partition the final $i$ elements into $k$ blocks. Thus
\begin{eqnarray*}
{n+1\brace k+1}=\sum_{i=k}^{n} \binom{n}{i}{i\brace k}.
\end{eqnarray*}