
I think it's very nice too.
My 2 cents...
Once I met problem “Derangements” on a math contest. The problem was to prove that f(n) always divisible by (n1). Here f(n)  number of derangements, n>1. I solved it in slightly different manner.
Proof:
Statement is obvious when n = 2, so let n > 2
Consider two different possibilities:
1. Derangements like
1 2 ... i ... n ....... n ... i
not hard to see that there are exactly (n1) * f(n2) of them
2. All other derangements, so
1 2 ... i ... n ....... j ... i j != n
Let's forget about last elements in both rows. If we look carefully, then we'll see that we can painlessly rename "i" in upper row to "n" since j!= n. And when we do that it becomes obviously that there are exactly (n1) * f(n1) such derangements.
So, f(n) = (n1) * ( f(n1) + f(n2) ) and is obviously divisible by (n1). 