Question
Find $\lim_{n\to \infty} P(n,n)$ where $$ P(n,k) = \sum_{j=0}^n (1-j){n\choose j} \frac{1}{(n-1)^j}\left( \frac{n-j}{n} \right)^k. $$
The origin of this sum is from this question. The probability that a box is empty after $k$ iterations is given by $\left( \frac{n-1}{n} \right)^n P(n,k)$. Verifications of this are also welcome, I just managed to guess it with numerical evidence. But that got me thinking about the limiting behavior.
Obviously for any particular $n$, if $k \to \infty$, only the term $j=0$ survives and we're left with just $P(n, \infty) = 1$. This means the probability is $\left( \frac{n-1}{n} \right)^n$ and that goes to $e^{-1}$. But what if $n$ goes to infinity with $k$? Let's consider the case $k=n$. More general approaches would be interesting too.
If we approximate (but are these allowed?)
$$ {n\choose j} \frac{1}{(n-1)^j} \approx \frac{1}{j!} $$
and (remember now $k=n$)
$$ \left( \frac{n-j}{n} \right)^n \approx e^{-j} $$
we get
$$ P(n,n) \approx \sum_{j=0}^n {\frac{1-j}{j!}}e^{-j} $$
and hence the limit would be
$$ \lim_{n\to \infty} P(n,n) = {\huge e^{\frac{1}{e}} \left(1-\frac{1}{e} \right)}. $$
And the probability would be
$$ e^{\frac{1}{e}-1} \left(1-\frac{1}{e} \right) = 0.335949071234. $$