Mersenne primes

Contrapositive. Suppose $n=ab$ with $1<a,b<n$. Then
$$2^n-1=2^{ab}-1=(2^a)^b-1=(2^a-1)(2^{a(b-1)}+\cdots+1).$$
Both factors are $>1$, so $2^n-1$ is composite.

Conclusion. If $2^n-1$ is prime, $n$ must be prime.

Example. $2^3-1=7$ prime, but $2^4-1=15$ composite.

Remark. The converse is false: $2^{11}-1=2047=23\cdot89$.