Show that $\frac{46!}{48}+1$ is not a power of 47

Suppose that $$ 47^n=\frac{46!}{48}+1. $$ Then, $7\mid 47^n-1$, so $47^n\equiv (-2)^n\equiv 1\pmod 7$. Hence, $6\mid n$. However, it means that $$ 47^6-1\mid 46! $$ but $47^2-47+1$ divides $47^6-1$ and $47^2-47+1=2163=21\cdot 103$. Thus, $2163\mid 46!$, but $103$ is prime and $103>46$, contradiction.

$$46!=48(47^k-1)$$

First since $47^{7}\equiv1\pmod{43}$ we know $7\vert k$. (Easy to see by brute force $7$ times)

Next since $47^{40}\equiv1\pmod{41}$ and $47^{2}\equiv35\neq1\pmod{41}$ and $47^{4}\equiv24\neq1\pmod{41}$ and $47^{5}\equiv26\neq1\pmod{41}$ we know at least one of $8$ or $10$ divides $k$.

Either case, $k\geq56$ which implies the right side is much larger than the left side, contradiction.