Find the remainder when $(34! + {75}^{37})^{39}$ is divided by $37$

Question

Since Fermat Theorem is $a^{36} \equiv 1 \mod {37}$, ${75}^{37}$ becomes ${75}^{36} \times 75$ and in $\!\!\mod {37}$ they both become $1$. I have $(34! + 1)^{39}$. I do the same again with $(34! + 1)^{36} (34! + 1)^3$ and the power of $36$ number also becomes $1$. I just dont know how to solve $(34! + 1)^3 \mod {37}$ at this point.

Answer 1

By Wilson $36!+1$ is divisible by 37.

Thus, we work with $$\left(\frac{-1}{-1\cdot(-2)}+1\right)^{3}.$$ I got the answer: $14$.