Sum of Stirling's number of first kind

Question

I am stuck at a proof

$\sum_{i=0}^ns(n,i) = n!$

Here $s(n,i)$ refers to sterling's number of first kind.

I can understand $s(n,n)$ refers to the permutation where all number are at their own position and $s(n,1)$ refers where each number is left shifted by 1

But i can not come up with a formal proof

Answer 1

This follows immediately from the definition.

The Stirling number of the first kind $s(n,i)$, also written $n\brack i$, is the number of permutations of $[n]=\{1,\ldots,n\}$ having exactly $i$ cycles. Every permutation of $[n]$ has some number of cycles, and clearly the maximum possible number of cycles is $n$ (for the identity permutation). How many permutations of $[n]$ are there altogether?