What is the remainder when $7!$ is divided by $23$?

(Note: $n!$ means "n factorial", defined as the product of all integers from $1$ to $n$.)
$7!$ is defined as $1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7$.

Note that $1\cdot 2\cdot 3\cdot 4 = 24 \equiv 1\pmod{23}$. So, $$7! \equiv 1\cdot 5\cdot 6\cdot 7\pmod{23}.$$Furthermore, we have $1\cdot 5\cdot 6 = 30\equiv 7 \pmod{23}$, so \begin{align*}
7! &\equiv 7\cdot 7 \\
&= 49 \\
&\equiv 3 \pmod{23}.
\end{align*}The remainder is $\boxed{3}$.