Probability that a random permutation has no fixed point among the first $k$ elements Is it true that $\frac1{n!} \int_0^\infty x^{n-k} (x-1)^k e^{-x}\,dx \approx e^{-k/n}$ when $k$ and $n$ are large integers with $k \le n$?
This quantity is the probability that a random permutation of $n$ elements does not fix any of the first $k$ elements.

Update: This argument only holds for some cases.  See italicized additions below.
Let $S_{n,k}$ denote the number of permutations in which the first $k$ elements are not fixed.  I published an expository paper on these numbers earlier this year.  See "Deranged Exams," (College Mathematics Journal, 41 (3): 197-202, 2010).  Aravind's formula is in the paper, as are several others involving $S_{n,k}$ and related numbers.
Theorem 7 (which I also mention in this recent math.SE question) is relevant to this question.  It's
$$S_{n+k,k} = \sum_{j=0}^n \binom{n}{j} D_{k+j},$$
where $D_n$ is the number of derangements on $n$ elements.  See the paper for a simple combinatorial proof of this.
Since $D_n$ grows as $n!$ via $D_n = \frac{n!}{e} + O(1)$ (see Wikipedia's page on the derangement numbers), and if $k$ is much larger than $n$,
the dominant terms in the probability
$\frac{S_{n+k,k}}{(n+k)!}$ are the $j = n$ and $j = n-1$ terms from the Theorem 7 expression.  Thus we have 
$$\frac{S_{n+k,k}}{(n+k)!} \approx \frac{D_{n+k} + n D_{n+k-1}}{(n+k)!} \approx \frac{1}{e}\left(1  + \frac{n}{n+k}\right) \approx e^{-1} e^{\frac{n}{n+k}} = e^\frac{-k}{n+k},$$
where the second-to-last step uses the first two terms in the Maclaurin series expansion for $e^x$.
Again, this argument holds only for (in my notation) $k$ much larger than $n$.
