Probability: k items assigned to n slot, find Pr. that at least 1 slot contains more than 1 item The following, rephrased problem is from the book "Introduction to Probability" on p.35.
I have no solution to this problem and I'm not sure if my solution is correct.
No homework, independent self-study!

Problem:
  The telephone numbers of k people are randomly and independently assigned to n locations with equal probability (one location can contain more than one number). What is the probability that at least one location contains more than one number?

My solution:
Solve the problem by computing the probability that exactly 0 and exactly 1 location(s) contain more than one number, respectively.


*

*number of possible outcomes: $n^k$

*number of favourable outcomes for exactly 0 location containing more than one number: $\binom{n}{k}k!= n \binom{n-1}{k-1}(k-1)!$

*number of favourable outcomes for exactly 1 location containing more than one number: $\sum_{\color{red}{i=2}}^{k} n \binom{n-1}{k-i}(k-i)!$


Thus, the result should be: $1 - \frac{1}{n^k} \sum_{\color{red}{i=1}}^{k} n \binom{n-1}{k-i}(k-i)!$
Here is some R code for the case $n=10$ and $k=5$, giving a probability of 0.639. Of course, that will blow up for large n, but not the matter here.
n <- 10
k <- 5
sum <- 0.0
for(i in 1:k) sum <- sum + n*choose(n-1, k-i)*factorial(k-i)
1.0 - sum/n^k
> 0.639

Question: Is my approach correct?
EDIT
As Henry said, I'm calculating the probability that at least two locations contain more than one number. Hence, the correct solution should be $1-\frac{1}{n^k} \binom{n}{k}*k!$, I guess.
 A: Your correction looks correct.
If you want to simulate this in R, the following will do so rather inefficiently
people <- 5
locations <- 10
samplesize <- 10000
atleast1 <- 0
set.seed(2014)
for (i in 1:samplesize){
   if(length(unique(sample(locations, people , replace=TRUE))) != people){
       atleast1 <- atleast1 + 1 
       }
   }
atleast1 / samplesize
1 - choose(locations, people) * factorial(people) / locations^people

giving 
> atleast1 / samplesize
[1] 0.7007
> 1 - choose(locations, people) * factorial(people) / locations^people
[1] 0.6976

which are reasonably close
A: Provided that both the "telephone numbers" and the "locations" are distinct, your solution (after Henry's remark) is absolutely correct. It is probably stated in the problem, but it must be true that $k \leq n$, else the probability is $1$. This is commonly known as the Birthday problem. You can read more if you're interested.
If the telephone numbers are considered not distinct, then the number of arrangements such that every location contains at most $1$ number (i.e. the complementary event) is $\displaystyle \binom{n}{k}$ and the total number of arrangements is $\displaystyle \binom{n + k - 1}{k - 1}$ (look at stars and bars), giving us a result of $\displaystyle 1 - \frac{\binom{n}{k}}{\binom{n + k - 1}{k - 1}}$.
