$25! \pmod {78125}$ $25! \pmod{78125}$ is a problem I'm working on.
Since $78125$ looked very divisible by $5$, I checked, and found that $78125 = 5^7$.
Then I thought, if there are seven factors 5 in $25!$, then $25! \equiv 0 \mod 78125$, but I only found $25$ to be divisible by 5 six times, so I don't think that got me anywhere.
Am I even on the right track here?
Thanks in advance for any help!

If $\frac{25!}{5^6}\equiv k \pmod{5}$, then $5\mid\frac{25!}{5^6}-k$, i.e. $5^7\mid 25!-k\cdot 5^6$ so $25!\equiv k\cdot 5^6 \pmod{5^7}$. What remains is to find $k$.
For that, note $\frac{25!}{5^6}=(1\cdot 2 \cdot 3 \cdot 4) \cdot (6
\cdot 7 \cdot 8 \cdot 9) \cdot (11 \cdot 12 \cdot 13 \cdot 14) \cdot (16 \cdot 17 \cdot 18 \cdot 19) \cdot (21 \cdot 22 \cdot 23 \cdot 24)\cdot(\frac55 \cdot \frac{10}{5} \cdot \frac{15}{5} \cdot \frac{20}{5})\equiv (4!)^6\equiv (-1)^6=1 \pmod 5$ , so $k=1$.
That gives us the answer: $25!\equiv 5^6 \pmod {5^7}$.
