Prove that for positive integer n we have $(2^n-1)^2 \mid 2^{n(2^n-1)} - 1$. How to solve this?
$$(2^n-1)^2 \mid 2^{n(2^n-1)} - 1$$

The expression $$\dfrac{(x+1)^x-1}{x^2}$$ is an integer for all positive integers $x$. We have $$(x+1)^x-1 = x^x+\binom{x}{1}x^{x-1}+\cdots+\binom{x}{x-1}x+1-1 \equiv 0 \pmod{x^2}.$$
