TITLE: Proof Binomial Coefficient Identity: $\sum_{k=0}^n\frac{k k!}{n^k}\binom{n}{k}=n$
QUESTION [15 upvotes]: Wolfram show that
$\displaystyle\sum_{k=0}^n\frac{k k!}{n^k}\binom{n}{k}=n$.
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How to prove this identity?
Thank you.

REPLY [4 votes]: Here's a way that is similar to Brian M. Scott but a slightly different take. We prove the equivalent identity
$$
\sum_{k=1}^{n} \frac{k k!}{n^{k+1}} \binom{n}{k} = 1.
$$
Suppose we sample $n+1$ times from a group of $n$ distinct objects. The pigeonhole principle tells us that the probability of picking some object twice is $1$. Now we also can think about the probability that our first duplicate pick will be on the $k+1$th selection where $k = 1, 2, \ldots n$. That is our first $k$ picks are distinct and the $k+1$st is one of the first $k$. Our first choice doesn't matter but we then must pick $k-1$ distinct objects -- this has probability $(n-1)(n-2) \ldots (n-k+1)/n^{k-1} = n!/(n^k (n-k)!)$ followed by a pick of one the first $k$ which has probability $k/n$. Since $k$ ranges over $1$ through $n$ the probability of getting a duplicate somwhere can also be written
$$ 
\sum_{k=1}^n \frac{kn!}{n^{k+1} (n-k)!} = \sum_{k=1}^n \frac{k k!}{n^{k+1}} \binom{n}{k} 
$$