You Don't See This Expression Every Day

the problem statement is equivalent to saying $x^x+x!\equiv 0\pmod{x+1}$ .

First, note that $x\equiv -1\pmod{x+1}$ . Thus, $x^x\equiv (-1)^x\pmod{x+1}$ . Thus, for all even numbers, $x^x\equiv 1\pmod{x+1}$ and for all odd numbers, $x^x\equiv -1\pmod{x+1}$ .

Now, let's look at $x!\pmod{x+1}$ . Note that if $x+1$ is composite, then $x!\equiv 0\pmod{x+1}$ , and when $x+1$ is prime, $x!\equiv -1\pmod{x+1}$ by Wilson's Theorem.

Thus, for $x^x+x!\equiv 0\pmod{x+1}$ , then $x!\equiv -1\pmod{x+1}$ and $x^x\equiv 1\pmod{x+1}$ . Thankfully, if $x+1$ is a prime number greater than $2$ , then $x$ is even, satisfying both conditions.

Now we must check $x=1$ explicitly because it is our special case. We have $\dfrac{1^1+1!}{1+1}=1$ which is an integer. So that means $x+1$ being any prime satisfies the condition.

Thus, we want $x+1$ to be the $100$ th prime. By the list, this is $541$ ; thus, $x=\boxed{540}$ .