Let $B_k$ denote the $k$ th Bernoulli number: $$B_0=1,\quad B_1=-\frac{1}{2},\quad B_2=\frac{1}{6},\quad B_3=0,\quad B_4=-\frac{1}{30},\ldots,\ B_{10}=\frac{5}{66},\ldots $$ In fact, $B_k=0$ for all odd $k\geq 3$ , so we will only consider $B_k$ for even $k$ . The following congruence is due to Ernst Eduard Kummer:

The interested reader should see also the congruence of Clausen and von Staudt for a similar result. As an example of Kummer's congruence, let $p=7$ and $k=4$ . Then: $$\frac{B_4}{4}=\frac{-\frac{1}{30}}{4}=-\frac{1}{120}\equiv 6 \mod 7$$ If we pick $h=10$ (so that $10\equiv 4 \mod (p-1)$ ) then: $$\frac{B_{10}}{10}=\frac{\frac{5}{66}}{10}=\frac{1}{132}\equiv 6 \mod 7$$ which is what the theorem predicted.