To be a bit more explicit: Your induction hypothesis is
For all $a\in\mathbb{R}_{\geq 0}$, if $n\geq a$ then $n!\geq a^n$.
You want to prove:
For all $b\in\mathbb{R}_{\geq 0}$, if $n+1\geq b$, then $(n+1)!\geq b^{n+1}$.
(I changed the letter to make it clearer; but of course that statement above is equivalent to one in which we have "$a$" instead of "$b$".
To prove the implication, you assume $n+1\geq b$. In order to apply the induction hypothesis, you would need $n\geq b$; if that holds, then your argument works: if $n\geq b$, then the induction hypothesis implies $n!\geq b^n$, and then multiplying by $n+1\geq b$ we get $(n+1)!\geq b^{n+1}$.
But what if we do not have $n\geq b$? The assumption $n+1\geq b$ does not guarantee this.
How could you find this out? Since you already know there is a counterexample, take your "inductive argument" out for a spin in the smallest counterexample. The smallest counterexample comes with $n=2$: assume $2\geq a$. Does it follow that $1!\geq a^1$? No; from $2\geq a$ you cannot conclude $1!\geq a$. In fact, if $2\geq a\gt \sqrt{2}$, you will have $a^2 \gt 2 = 2!$; the problem being that you don't get to apply that induction hypothesis on any $a$ greater than $1$ (and the conclusion fails for any $a$ greater than $\sqrt{2}$).