Why doesn't $0! = 1$ in the context of this general term?

Question

Is my instructor wrong to say that $\left\{0,\frac{1!}{4},\frac{2!}{9},\frac{3!}{16},\dots\right\} = \left\{\frac{(n-1)!}{n^2}\right\}$?

My understanding is that at $n=1$, $\frac{(n-1)!}{n^2}$ should equal $1$, not $0$.

Answer 1

The problem isn't that $0!=1$, the problem is that the first term in the sequence doesn't match the pattern of the other terms. The sequence whose $n^{\text{th}}$ term is $\dfrac{(n-1)!}{n^2}$ looks like $$1,\ \dfrac{1}{4},\ \dfrac{2}{9},\ \dfrac{6}{16},\ \dots$$ whereas your sequence starts with $0$, not $1$.