The classification of finite simple groups was one of the great mathematical achievements of the 20th Century. It is also one where a single result on the order of the groups played a key role, namely the Feit–Thompson theorem, or odd order theorem:
Theorem. (Feit-Thompson, 1963) Every group of odd order is soluble*.
The proof is famously long, at 255 pages, and has recently been Coq-verified [1].
The derived subgroup of a soluble group is a proper normal subgroup, and so a soluble group is simple only if it is abelian. Therefore, the Feit-Thompson theorem has the following corollary:
Corollary. Every non-cyclic finite simple group has even order.
There are other results in this vein, with much shorter proofs. For example, Burnside's theorem (Wikipedia contains a proof):
Theorem. (Burnside, 1904) Let $p, q, a, b\in\mathbb{N}$ with $p, q$ primes. Then every group of order $p^aq^b$ is soluble.
Therefore, every non-cyclic finite simple group must have order divisible by three primes. Moreover, at least one of these primes occurs twice in the prime decomposition of the order:
Theorem. (Frobenius, 1893) Groups of square-free order are soluble.
You can find a proof of this theorem on Math.SE here. The answer there links to the article [2], where the theorem is Proposition 17 (page 9). The article also claims that the result is due to Frobenius in [3].
*In American English, solvable.
[1] Gonthier, Georges, et al. "A machine-checked proof of the odd order theorem." International Conference on Interactive Theorem Proving. Springer, Berlin, Heidelberg, 2013.
[2] Ganev, Iordan. "Groups of a Square-Free Order." Rose-Hulman Undergraduate Mathematics Journal 11.1 (2010): 7 (link)
[3] Frobenius, F. G. "Uber auflösbare Gruppen." Sitzungsberichte der Akademie der Wiss. zu Berlin (1893): 337-345.