Wolstenholme's theorem that the numerator of H_{p-1} is a multiple of p^2 is a bit tricky to prove, but the weaker result that the numerator is a multiple of p

Wolstenholme's theorem that the numerator of H_{p-1} is a multiple of p^2 is a bit tricky to prove, but the weaker result that the numerator is a multiple of p is very straightforward. Here are two ways. ONE Multiply through by P := 1.2.3...(p-1); note that if a fraction's numerator is a multiple of p after doing this, it must have been before as well. Now what we have is obviously an integer; it's P/1 + P/2 + P/3 + ... + P/(p-1). Clearly none of these terms is a multiple of p. And no two of these terms can be equal mod p, because if P/a-P/b is a multiple of p then so is ab(P/a-P/b) = (b-a)P, but that's impossible: neither b-a nor P is a multiple of p, and p is prime. So their values mod p must be 1,2,3,...,p-1 in some order, and when you add those up you get a multiple of p. We're done. (For an audience familiar with the relevant ideas, this can be considerably abbreviated: "Mod p, this is the sum of the inverses of 1,2,...,p-1, and these are just 1,2,...,p-1 in some order, whose sum is 0, QED.") TWO Group the fractions in pairs: 1/k and 1/(p-k). The sum of each pair is p/k(p-k) and of course when we add these up the numerator is still a multiple of p (because there are no p's in the denominator). Done. (I think the first of these is really the simpler and more informative of the two, even though the second is shorter.)