Suppose that $L$ is regular. Then the language $P = 1^+ - L$ is also regular. Observe that $P = \{1^p \mid p \text{ is prime } \}$. Since $P$ is infinite, the pumping lemma can be applied: there exists $N$ such that every word $w$ of $P$ of length at least $N$ can be written as $w = xyz$, where $x, y, z \in 1^*$, $0 < |y| \leqslant N$ and $xy^*z \subseteq P$.
Let us apply this result to $w= 1^p$, where $p$ is a prime number $> N+1$. Let $n = |xz|$. Since $|w| = |xyz| > N+1$ and $|y| \leqslant N$, one has $n \geqslant 2$. Since $xy^*z \subseteq P$, one has $xy^nz \in P$. However $$|xy^nz| = |xz| + |y^n| = n + n|y| = n(1+|y|).$$Since $n(1+|y|)$ is a composite number, one gets $xy^nz \notin P$, a contradiction.