For $ n = 1$ this is obvious, so assume by induction it is true for $ n-1$. We make piles of sum $ mn$, $ m < n$. As long as there are at least $ n$ cards $ a_{1},...,a_{n}$, either the set $ \{a_{1},a_{1}+a_{2},a_{1}+a_{2}+a_{3},...\}$ contains a multiple of $ n$, or - by pigeonhole - two have the same remainder $ \mod n$, so their differences form a subset with sum divisible by $ n$. So we have such a set When there are less than $ n$ left, the remaining cards form such a set Applying the induction hypothesis with the values for $ m$ as card numbers we have $ k$ piles of $ n(n-1)! = n!$ as desired.

Hello. I understand the overall solution, but I do not understand what do you mean exactly by by aplying the values of m as card numbers.