A permutation and combination question (plus summation of a series) A student is browsing in a second-hand bookshop and finds $n$ books of
interest. The shop has $m$ copies of each of these $n$ books. Assuming he never
wants duplicate copies of any book, and that he selects at least one book, how
many ways can he make a selection? For example, if there is one book of interest
with two copies, then he can make a selection in 2 ways. 
So how I'm solving this is as follows: If he has to chose 1 book, he can do hat in $m$ ways. If he has to chose 2 books, he can do it in $m^2$ ways. Simialrly he choses $n$ books in $m^n$ ways. The total number of ways he can make this selection is $$ \sum m + m^2 +..... m^n$$ which is a geometric series with sum $$\frac {(m) (1-m^n)} {1-m}$$ The answer however is apparently $(m+1)^n -1$ . Where am I going wrong?

For each kind of books, the person can choose $m$ copies, plus not to choose any of the kind. Since there are $n$ kinds, there are $(m+1)^n$ outcomes. Because the person chooses at least 1 book, this excludes the outcome that none are chosen, so there are $(m+1)^n - 1$ outcomes.
