Let's represent each number \(13^m\) as \(13^m = q_m2013 + r_m\).
There are 2049 natural numbers \(\le 2048\). So there are 2049 numbers of the form \(13^m\). However, there are only 2013 remainders when you divide numbers by 2013. So two of the \(13^m\) must have the same remainder mod 2013.
Let's choose a and b as the names of two numbers such that \(13^a\) and \(13^b\) have the same remainder mod 2013. Let's call this shared remainder R. That is, we have \(13^a = q_a2013 + R\) and \(13^b = q_b2013 + R\).
But then \(13^a - 13^b = q_a2013 - q_b2013 = (q_a - q_b)2013\). Since \(q_a\) and \(q_b\) are integers, \(q_a -q_b\) is an integer. So this means \(13^a - 13^b\) is divisible by 2013, which is what we needed to show.