If $n$ is a composite number, let it be the product of $ab$ where integers $a$ and $b$ satisfy $1 < a, b < n$. We then have $$2^n-1 = \Big(2^a\Big)^b-1 = (2^a -1)\Big((2^a)^{b-1}+(2^a)^{b-2}+ \cdots + 1\Big)$$ Note $1 < a < n \implies 1 < 2^a -1 < n$. This means $2^n -1 $ is not a prime number which is a contradiction!