Alright bump time - time for me to spoil this problem.
So, the sum from n = 0 to N of 1/n! is always < e for all N. Now, 1/(N * N!) * sum(1/n!) - e (where sum(1/n!) is from n = 0 to N) = 1/(N * N!) - sum(1/n!, n > N) = 1/N! * (1/N - 1/(N+1) - 1/((N + 1)(N + 2)) - ...) = 1/N! * (1/(N * (N + 1)) - 1/((N + 1)(N + 2)) - 1/((N + 1)(N + 2)(N + 3) - ...), which can be shown to be > 0.
Now, the fun part is suppose e is rational. Then e = p/q, where p is an integer and q is a natural number. Let N = q
Then the sum from n = 0 to q of 1/n! < p/q < sum from n = 0 to q of 1/n! + 1/(q * q!)
Then 0 < p/q - sum from n = 0 to q of 1/n! < 1/(q * q!)
Then 0 < p * q! - (q + q * (q - 1) + ... + q!) < 1
However, the middle part is an integer, so that's a contradiction.
Next question:
(AIME 2001 Version II Problem 3) The sequence a1, a2, a3, ... is defined by a1 = 211, a2 = 375, a3 = 420, a4 = 523, an = an-1 - an-2 + an-3 - an-4. Find a531 + a753 + a975.