Q:

Ways to get a solution of this equation

The equation is
$x!+(x-1)!+(x-2)!+...+1!+0!=x^x+x^{x-1}+x^{x-2}+...x^1+x^0$
I got a solution (1) by substituting $1$ in $x$
I want to know if there is another way to get the solution and also another solution to this equation. 

A:

If you only care about natural numbers, then:
$x! = x \times (x-1) \times \ldots \times 2 \times 1 < x \times x \times \ldots \times x \times x = x^x$ when $x > 1$. Similarly for all the terms you're adding up, so the only possible solutions are 0 and 1, and 0 only works if you use the commonly accepted convention that $0^0 = 0! = 1$.

