Smallest Guaranteed hash collision cycle length If I take the sha-256 of an empty string, and apply the hash function $2^{256}!$ times, will I end up with the same hash that I started with?

Is the smallest required cycle equal to the LCM of $1$ to $2^{256}$?
In a uniform random function, which is a mostly reasonable model for SHA-256, you are not guaranteed that any particular element lies in a cycle: it could go $A \mapsto B \mapsto C \mapsto B$ and therefore never return to $A$.

In fact, only a tiny minority of elements are likely on cycles at all—the expected number of elements on cycles in a uniform function of $n$ elements grows with $\frac 1 2 \sqrt{2\pi n}$; for SHA-256, this is about $2^{128}$ of $2^{256}$ possible 256-bitstrings, i.e. the probability that any particular element is on a SHA-256 cycle is about $2^{-128}$.

However, if you just want to know how long you must pursue a hash chain before it cycles somewhere—not necessarily back to where you started ($A$), but somewhere you started going in circles lost in the desert trying to follow your own tracks like a pair of Belgian detectives ($B$)—the length of that chain turns out to have the same distribution as the number of elements on a cycle, with expectation $\frac 1 2 \sqrt{2\pi n}$, and the expected length of the cycle itself is $\frac 1 4 \sqrt{2\pi n}$.

So on average you'd have to follow $2^{128}$ elements before you ever come back upon your footsteps somewhere in a SHA-256 hash chain, and the average cycle length is about $2^{127}$, so you'd skip half your footsteps when you cycle back.

Here's an illustration that I stole from fgrieu of a 7-bit hash function—the vast majority of points, the grey dots, are not on cycles at all; only the red dots are on cycles:



Of course, in principle it could turn out that SHA-256 is actually a permutation of $\{0,1\}^{256}$ and is a single giant cycle, in which case you would actually have to tread $2^{256}$ steps—this, rather than the much larger $2^{256}!$, is the maximum number of steps before you are 100% guaranteed to have cycled back upon your footsteps.  But it is exceedingly unlikely that SHA-256 is a permutation, let alone a cyclic permutation—this would be an astonishing development if proven.

See Harris 1960 (paywall-free) for everything you need to know about statistics of uniform random functions, permutations, and derangements.