Why is it that for $n<60$, the simple groups are precisely the cyclic groups $\mathbb{Z}_N$ for prime $n$? I was reading this Wikipedia article and came across this

For groups of order $n<60$, the simple groups are precisely the cyclic groups $\mathbb{Z}_n$ for prime $n$.

I was wondering, why is this true for $n<60$? What is so special about this value?
 A: Well if the order of a group is divisible by at most two primes then it is either a cyclic group of prime order or has a proper normal subgroup (or the trivial group of order $1$). So any non-abelian finite simple group is divisible by at least three primes (see Burnside's Theorem)
The first possibility with at least three primes is $30=2\times 3 \times 5$ - but if this were simple, Sylow's theorems would tell us that there would be six subgroups of order $5$ containing $24$ elements of order $5$ and ten subgroups of order $3$ containing $20$ elements of order $3$. So that doesn't work.
Then [added because my arithmetic is poor, in response top a comment] there is $42=2\times 3\times 7$ which Sylow tells us has a normal subgroup of order $7$. 
The next possibility is $60$ which does work.
The Sylow Theorems can be awkward to use in general (for example they don't provide any easy proof of Burnside's Theorem), but they can tell us a lot about groups with simple structures. For example, if we have a factor $3$ in the order of the group we also need one of $4,7,10,13,16 \dots$. A factor of five comes with a factor from $6, 11, 16, 21 \dots$, and seven comes with $8, 15, 22 \dots $
Now suppose, for example, we have $4$ Sylow $3$ subgroups in our simple group $G$. Then $G$ acts transitively by conjugation on those subgroups, and this action gives a homomorphism from $G$ onto a non-trivial subgroup of $S_4$. Since $G$ is simple, this must have trivial Kernel and $G$ is isomorphic to a subgroup of $S_4$. So four subgroups won't do.
If we have a factor seven, we need at least $8$ as another factor, plus another prime. If we take $3$ we get $168$ which is the order of the next non-abelian simple group after order $60$.
Daniel Fisher's comment that smaller groups "aren't complicated enough to be simple" is spot on.
A: All groups of order smaller than $60$ are solvable. And the simple groups that are solvable are precisely those of prime order.
A: Note that 60 is the least non square free product of 3 powers of distinct primes(= $2^2\bullet3\bullet5$), all non-negative integers less than 60 is one of the following cases and any group of this order is either abelian or not simple or both:
p-group : Either cyclic (abelian) or not-simple (By the 1st Sylow)
$p^a q^b$ : Burnside's $p^a q^b$ theorem states they are not simple.
pqr : Assume some group of order pqr is simple and p<q<r then calculate the smallest numbers can be $n_p, n_q, n_r$, then count the smallest number of elements of Sylow p,q,r-subgroups reveals that the sum is exceed pqr which is a contradiction.
0 : $<e>$, which is abelian and not simple since everything is equal to e, and has no proper normal subgroup.
