the argument that P(k) -->P(k+1) uses the fact that there are at least 2 coins in the box ("...and remove a different coin...."). so the proof is not complete unless we can prove P(2). but P(2) is clearly false, since we might have two different denominations in the box.
to put it another way, the proof shows that if we form two different subsets of k coins, A and B, from a set of k+1 coins, and each has all coins of the same denomination, then the "base case" is actually where removing one coin leaves one coin. that is, we have two coins. and the base case fails.