Is 100! + 1 a Prime Number
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Here's one I've found on Quora, but let ChatGPT solve it there.

100! is such a big number that the answer to the
question is "probably no".

If it appears as an exam, it is probebly the second part if a
question, and the first is:

  Prove Wilson's theorem.

Wilson's theorem states that.

  For any natural number n>1:
  n is a prime number iff and only if (n-1)!≡-1 mod n

Both directions can be proven easily:

First direction:
Suppose that n is a prime numbers. Then
each number m in 1,2,...,n-1 has a reciprocal k in
1,2,...,n-1 such that

  mk!≡1 mod n
  
because n is a prime number. You can use Fermat's little
theorem to find the reciprocal.
Now, up to two numbers in 1,2,...n-1 are their own reciprocals:
1 and n-1. The proof is that if
A,B are in 1,2...n-1 and n divides A²-B²
Then

  n | (A - B)(A + B)

and because n is a prime number n divides either A-B or A+B
In other words
  A = B or (n-A)=B

in particular, when A² ≡ B² ≡ 1 mod n

For n=2, there is nothing to prove. Otherwise, each of the numbers
2,3,...,n-2 can be paired with its reciprocal, thus

  (n - 2)! ≡ 1 mod n ⇒ (n - 1)!=(n - 1)(n - 2)! ≡ -1 mod n

Second direction:
Suppose (n - 1)! ≡-1 mod n, but n is composite. Then n has a divisor
d which is other from 1 and n. Thus,
n is in 2,3,...,n-1. Thus,
d | (n - 1)!

But

  (n - 1)! ≡ -1 mod n

means that there exists an integer k such that:

  (n - 1)! = kn -1 ⇒ (n - 1)! + 1 = kn

and d divides n, thus

  d | (n - 1)! + 1    (1)

d also divides (n - 1)!

and with (1):

  d | 1  = d = 1

which contradicts our hypothesis that n is composite.
Thus, n is a prime number, and Wi;son's theorem is proven!

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Let us now prove using Wilson's theorem that 100!+1 is composite:

101 is a prime number, thus, according to Wilson't theorem:

  100! ≡ -1 mod 101

which means that there exists an integr k, such that:

  100! = 101k - 1 ⇒ 100! + 1 = 101k

and k>1 because 100!>101 ⇒ k>1

Thus,

  100! + 1 = 101k

is composite.