I've been playing around with Stirling Numbers and have stumbled upon the following relation which I think is true (I've tested for $n=3,4,5,6$). I can't seem to prove this however, and was wondering how one could prove the following equality.
Let $\begin{bmatrix}n \\ k\end{bmatrix}$ denote the Stirling number of the first kind for $n$ elements and $k$ cycles. Prove that $$\begin{bmatrix}n+1 \\ 2\end{bmatrix}=\sum_{k=1}^{n}k\begin{bmatrix}n \\ k\end{bmatrix}$$