We can prove something nearly as good: for positive integers x, x² divides f(x):=x^{x+1}+(x+1)^x -1. Since x+1≥2, it's immediately obvious x² divides x^{x+1}. Moreover, when we expand (x+1)^x with the binomial expansion theorem, we'll get a whole bunch of terms divisible by x², plus the remaining terms xC1 * x + xC0 * 1, which simplifies to x²+1. So when we finally take off 1, we'll get something divisible by x². Furthermore, by considering the cases x=0,1,2,3 mod 4 separately, for x>1 we deduce that f(x) is divisible by 4. Therefore, _if x is odd and >1_ we can conclude f(x) is divisible by 4x²=(2x)², as required. But if x is even, this needn't hold, a another commenter has already shown - we can only deduce divisibility by x².