Remainder of a combination Problem from a contest:
What is the remainder when $\binom{169}{13}$ is divided by $13^5$?
I thought that Wolstenholme's/Babbage's would help, but not entirely sure how.

$$\binom{169}{13}=13\cdot\frac{168\cdot \ldots \cdot 157}{12!}=13\cdot\frac{(13^2-1)\cdot\ldots\cdot(13^2-12)}{12!}$$
$$\frac{1}{12!}\prod_{k=1}^{12}(13^2-k)=\prod_{k=1}^{12}\left(1-\frac{13^2}{k}\right) = 1-13^2 H_{12}+C\cdot13^4$$
and since $H_{12}\equiv 0\pmod{13^2}$ by Wolstenholme's theorem, it follows that:
$$\frac{(13^2-1)\cdot\ldots\cdot(13^2-12)}{12!}\equiv 1\pmod{13^4}$$
and so:
$$\binom{169}{13}\equiv 13\pmod{13^5}.$$
