Why is it that for $n<60$, the simple groups are precisely the cyclic groups $\mathbb{Z}_N$ for prime $n$?

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.