Really i'm interesting to know if there is any Theorem show that always there exist an integer $k>1$ for which :$\displaystyle n! \bmod {n}^{k}\neq 0$? for example we take this as example : $n=10 $ with $k=2$ to $100$, we see that from $k=3$ to $100$ :$\displaystyle 10! \bmod {10}^{k}\neq 0$.
My modest question is: Is there any theorem or any known result show that always there exist an integer $k>1$ with $k<n$ for which :$\displaystyle n! \bmod {n}^{k}\neq 0$?.
Thank you for any help
If $n! \equiv 0 \pmod {n^k}$, then $n^k \mid n!$. To find a small $k$ such that $n^k \nmid n!$, we consider the largest prime $p$ dividing $n$. If $p^k \nmid n!$, then clearly $n^k \nmid n!$.
Fortunately, we know exactly how many times $p$ divides $n!$.
So choosing any $k$ satisfying $$ k > \sum_{j \geq 1} \left \lfloor \frac{n}{p^j} \right \rfloor$$
will suffice. Note that $$ \sum_{j \geq 1} \left \lfloor \frac{n}{p^j} \right \rfloor \leq \sum_{j \geq 1} \frac{n}{p^j} = \frac{n}{p} \frac{1}{1-\frac{1}{p}} = \frac{n}{p-1},$$ which is always less than $n$ as long as $n$ has a prime factor larger than $2$.
The case when $n = 2^m$ for some $m$ is a bit more annoying and should be handled separately. But having only one prime to worry about simplifies the analysis.
