How does one prove that the # of n- element permutations with k left-right minima is given by the 1st kind stirling number?
I understand one should consider sigma(1)=n and sigma(1)=/=n, but I am totally lost on the process A left-right minimum of a permutation σ is an index σ(j) such that σ(j) < σ(i) for all i < j
Think of a permutation as a list of the numbers $1,\ldots,n$ in some order. One gets such a permutation by "inserting" $n$ somewhere in a permutation of $1,\ldots,n-1$. If one inserts it at the beginning one gains one left-right minimum. If one inserts it elsewhere the number of left-right minima remains the same.
So to get an $n$-permutation with $k$ left-right minima, either one prefixes an $(n-1)$-permutation with $k-1$ minima with an $n$, or else inserts an $n$ into an $(n-1)$-permutation with $k$ minima in one of the $n-1$ possible locations that's not at the front.
This gives the Stirling recurrence.