Which explanation correctly justifies why $0!=1$ using the recursive identity $n!=n\cdot (n-1)!$?

Define $0!$ as the average of $1!$ and $(-1)!$

Use the sum formula $0!=1+0$, so $0!$ must be $1$

Set $n=1$ in $n!=n\cdot (n-1)!$ to get $1!=1\cdot 0!$. Since $1!=1$, it follows that $0!=1$.

Since factorials count arrangements, and there is one way to arrange zero objects, it follows that $0!=1$