The expected number of times to get the same ball from $n$ different balls This looks very simple, but it's totally over my head. Any clue is appreciated.

Randomly choose one ball out of $n$ different balls, calculate the expected number of experiments when the same ball appears again. 

My solution is: $E(e)= 2P(n=2) + 3P(n=3) + \cdots + dP(n=d) + (d+1)P(n=d+1)$
The ball will be the same to the one of the previous balls in the $d+1$ experiments for sure. The probability:
$$P(e=i) = (i-1)d(d-1) \cdots (d-i+1)/d^i,$$
but I don't know how to find $E(e)$. Any suggestions would be appreciated.
 A: I think that you are on the right track.
Two observations:
1. There must be 2 or more trials
2. If there are n+1 trials then one ball will be chosen twice.
Now, let p(m) be the probability that a ball is chosen twice in m trials. This is equivalent to the probability that there are m different balls in the m trials. So, 
p(m) = 1 - p(m different balls)
     = 1 - nPm/(n^m)
Now, E = sum i*p(i), i=2...n
       = sum i*(1 - nPm/(n^m)), i=2..n
im sure it is possible to simplify this expression further...
A: It should be:
$$P(e=i)=\frac{(i-1)d(d-1)(d-2)\ldots (d-i+2)}{d^i}$$
I don't think there is a nice solution for calculating the expected value. Numerically, for $d=1,2,3,4,5,6\ldots$, I get the expected values are $2,\frac{5}{2},\frac{26}{9},\frac{103}{32},\frac{2194}{625},\frac{1223}{324}, \frac{472730}{117649}$.
The numerator is identified by OEIS as the integral
$$\int_0^\infty e^{-x}(1+x/d)^ddx$$
while the denominator is identified by OEIS the numerator (upon reducing to lowest terms) of the following expression:
$$\frac{(d+1)^d}{d!}$$
