The pebble sequence Let we have $n\cdot(n+1)/2$ stones grouped by piles. We can pick up 1 stone from each pile and put them as a new pile. Show that after doing it some times we will get the following piles: $1, 2, \ldots n$ stones.
Example: $n = 3$
Let we have 2 piles with 3 stones of each.
$$3 3 \to 2 2 2 \to 1 1 1 3 \to  2 4 \to 1 3 2$$

This was originally proved by Jørgen Brandt in Cycles of Partitions, Proceedings of the American Mathematical Society, Vol. 85, No. 3 (Jul., 1982), pp. 483-486, which is freely available here. The proof of this result covers the first page and a half and is pretty terse.
First note that there are only finitely many possible sets of pile sizes, so at some point a position must repeat a previous position. Say that $P_{m+p}=P_m$ for some $p>0$, where $P_n$ is the $n$-th position (set of pile sizes) in the game. It’s not hard to see that the game will then cycle through the positions $P_{m+k}$, $k=0,\dots,p-1$, repeatedly. The proof consists in showing that when there are $\frac12n(n+1)$ pebbles, the only possible cycle is the one of length one from $\{1,2,\dots,n\}$ to $\{1,2,\dots,n\}$.
Brian Hopkins, 30 Years of Bulgarian Solitaire, The College Mathematics Journal, Vol. 43, No. 2, March 2012, pp. 135-140, has references to other published proofs; one appears to be to a less accessible version of the paper Karatsuba Solitaire (Therese A. Hart, Gabriel Khan, Mizan R. Khan) that MJD found at arXiv.org.
Added: And having now read the Hart, Khan, & Khan paper, I agree with MJD: the argument is quite simple, and it’s presented very well.
