Let $p_{n,k}$ be the probability that a random permutation from $S_n$, the symmetric group of order $n!$, has exactly $k$ fixed points. I am trying to compute $\lim _{n\to \infty}p_{n,k}$. After playing with it a bit, I am relatively confident the limit is $\frac{1}{k!}$, although I have yet to come up with a proof.
Here's what I've tried so far. If $a_{n,k}$ is the number of elements of $S_n$ with $k$ fixed points, then we have that
$ a_{n,k}=\binom{n}{k}a_{n-k,0} $
That is, there are $\binom{n}{k}$ ways to choose $k$ fixed points, and after those are chosen, there are $a_{n-k,0}$ to permute the remaining $n-k$ elements with $0$ fixed points. After computing $a_{n,0}$ by hand for small values of $n$, I was able to look up the recursion formula $a_n=na_{n-1}+(-1)^n$ from oeis.org. Using this, I was able to come up with my "guess" of $\frac{1}{k!}$, but this direction doesn't seem to be leading me towards finding a proof.
Any ideas how to proceed? Group theory is not my strong point, so I would not be surprised if there is a result that makes this problem almost trivial. A pointer to any such results would be excellent.