What makes a number a perfect power?

Let there be $n > 2$ $M$ s on each side of the equation. Then $nM = M^n$ , which solves to $M = n^{\frac{1}{n - 1}}$ for positive real solutions of $M$ .

Since $n < 2^{n - 1}$ for $n > 2$ , we have $M = n^{\frac{1}{n - 1}} < (2^{n - 1})^{\frac{1}{n - 1}}$ or $M < 2$ for $n > 2$ .

This means that if a positive integer $M$ exists such that $n > 2$ , $M < 2$ , which means $M = 1$ . However, if $M = 1$ , then $nM = M^n$ would be $n \cdot 1 = 1^n$ , or $n = 1$ , which cannot be true as $n > 2$ .

Therefore, there are no positive integers $M$ such that $n > 2$ .

There is no answer to satisfy such equations except 0 and 2. Since the question states n to be greater than 2, no solution exist