Simplify $26! \cdot x \equiv 730^{171} \pmod{232}$ I want to simplify the expression 
\begin{align}
26! \cdot x \equiv 730^{171} \pmod{232}
\end{align}
My first step is to solve for $26! \cdot a \equiv 1 \pmod{232}$, that is, find the inverse of $26!$ modulo $232$.
To simplify, I should first calculate $26! \bmod{232}$. Using Wilson's theorem seems to be the best approach. But since $232 = 2^3 \cdot 29$ is not a prime number, I have to split up the expression using the Chinese Remainder Theorem:
\begin{align*}
\begin{cases}
26! \equiv a \pmod{29}\\
26! \equiv b \pmod{8}
\end{cases}
\end{align*}
I can use Wilsons theorem to solve the first equation as follows:
\begin{align*}
28! = 26!\cdot 27 \cdot 28 \equiv 26!(-2)(-1) \equiv 26!\cdot 2 \equiv -1 \pmod{29}
\end{align*}
Using that $2\cdot 15 \equiv 1 \pmod{29}$, we find
\begin{align*}
26! \equiv -15 \equiv 14 \pmod{29}
\end{align*}
However, I cannot use Wilsons theorem to solve the second equation. Any suggestions on how to solve that? Or might there be a better way to calculate $26! \bmod{232}$ altoghether?

Hint: Divide by $2^3$ on both sides (and the modulus) to obtain the equivalent congruence: $$(26)(25)(3)(23!) x \equiv (365)^3(730)^{168} \pmod{29}$$
Note that the left side contains $(26)(25)(3)(23!)$, which is relatively prime to $29$ and thus invertible modulo $29$ (like in the work you have done, Wilson's Theorem can be useful here). Then note that $730$ and $29$ are relatively prime; this permits you to use Fermat's Little Theorem (or Euler's Theorem) to simplify $730^{168}$.
It should be more straightforward from there.
