Remainder of $98!$ modulo $101$ My question is:

What would be the remainder when $98!$ would be divided by $101$?

Though this question is very easy but I'm a little confused about my concepts.
I have found multiples of $2$ and $7$ in $98!$, thus eventually I found that it contains $14^{10}$ which is divisible by 101,so the answer would be $0$ as the remainder.
Is my answer correct?
 A: $101$ is prime. It's a well-known fact that $(p-2)!\equiv1$ mod $p$. (The proof is that mod $p$, every nonzero number has a multiplicative inverse to pair with, except $1$ and $p-1$ which are each their own inverse.) So $99!\equiv1$ mod $101$. Find the multiplicative inverse of $99$, and multiply by it.
A: We have $(98!)(99)(100)=100!$. Thus by Wilson's Theorem,
$$(98!)(-2)(-1)\equiv -1\pmod{101}.$$
Let $x=98!$. Then $2x\equiv -1\pmod{101}$. It follows that $2x\equiv 100\pmod{101}$ and therefore $x\equiv 50\pmod{101}$.
A: Although the solutions here are outstanding, I want to share with you another approach.
Note that $98!\cdot99\cdot100\equiv(-1)(mod 101)$ from Wilson`s Theorem, we can get rid of the $100$ in the Left Hand Side of the equation (by modular arithmetic) and use the Fermat Little Theorem, as follows:
$99^{101-1} = 99\cdot99^{99}\equiv1(mod101)$
Using the fact that the inverse of $99$ is $99^{99}(mod101)$, we can begin our calculation from $99^{99}$ as follows:
$99^{99}=99\cdot(99^2)^{49} = 99\cdot4^{49}=99\cdot(4^7)^7=99\cdot22^7=99\cdot22\cdot22^6=99\cdot22\cdot(22^3)^2=99\cdot22\cdot43^2=57\cdot31\equiv50(mod101)$
Note that Modular Arithmetic is widely used here.
A: Hint By Wilson's Theorem, $$100! \equiv -1 \pmod {101} ;$$ on the other hand, $$100! = 100 \cdot 99 \cdot 98! .$$
