Let the roots be $r_1, \dots, r_n$. Then we want to show that $\gcd(r_1 \dots r_n, \sum_{i=1}^n \prod_{j \ne i} r_j) = 1$. This follows because $\gcd(r_k, \sum_{i=1}^n \prod_{j \ne i} r_j) = \gcd(r_k, \sum_{i=1}^n \prod_{j \ne k} r_j) = 1$.

Let $s_1, \dots, s_n$ are roots of $P(x)$ then we can write $P(x)$ like this $\prod_{(i=1)}^{n}(x-s_i)$ it is easy to show that $\gcd(s_1, \dots, s_n , a_0) = 1$ $\Rightarrow$ $\gcd(\prod_{(i=1)}^{n}s_i , a_0 ) = 1$ then $\gcd(a_1 , a_0) = 1$

Vieta's relations, and use the fact that each root is coprime to $a_1$.