Good evening, I'm trying to find an asymptotic of this sum:


$$\sum\_{j=0}^n (-1)^j {n \choose j} (n - j)^n = n^n - {n \choose 1} (n - 1)^n + {n \choose 2} (n - 2)^n + ... + (-1)^n {n \choose n} (n - n)^n $$


I think there is no close form. But I don't know how to calculate an asymptotics than. Maybe I should try to approximate it by some integral?


Thank you for any help!



Asymptotic of a sum involving binomial coefficients
Separate the different roles that $n$ plays in this sum, and look at
$$\sum\_{j=0}^n (-1)^j \binom{n}{j} (x-j)^n.$$
If $f(x)=f\_n x^n + (\mbox{lower order terms})$ is any polynomial of degree $n$, then $\sum\_{j=0}^n (-1)^j \binom{n}{j} f(x-j)$ is $n! f\_n$. So your sum is equal to $n!$, and asymptotics are given by Stirling's formula.


