The theorem that $\binom {n} {k} = \frac {n!} {k! (n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0 <k < n$. A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately. We treat binomial coefficients like $\binom {5} {6}$ separately already; the theorem assumes ...
What I would say is that you can multiply any non-zero number by infinity and get either infinity or negative infinity as long as it isn't used in any mathematical proof. Because multiplying by infinity is the equivalent of dividing by 0. When you allow things like that in proofs you end up with nonsense like 1 = 0. Multiplying 0 by infinity is the equivalent of 0/0 which is undefined.
So what IS the Holy Bible / The Great Standardization Document of All Definitions for Mathematics? Because people are often fighting over different definitions of mathematical entities, 0 being one of such examples (French always start a flamewar when someone says 0 is not positive, because for French, 0 is positive and negative at the same time :P ). Same goes with definitions of angles, or ...
Your numerical observations are correct and align precisely with the heuristic framework derived from Artin’s primitive root conjecture for algebraic numbers, combined with Chebotarev density and Kummer theory. Let $\alpha = (1+\sqrt {5})/2$. The Pisano period $\pi (p)$ is essentially determined by the multiplicative order of $\alpha$ (or $-\alpha^2$) in the appropriate finite field. 1 ...