To do it inductively what you can do is to use a generalized form of the induction principle:
Suppose that $P(x_1)$ is true and that for all $n$ that $(\forall j< n: P(x_j)) \Rightarrow P(x_n)$, then $P(x_n)$ is true for all $n>0$.
The difference here is that we instead of using the immediately previous number in the induction step allow us to use all previous ones(*).
Obviously the statement is true for $1$.
Now suppose it's true for all numbers below $q$. As we've seen that it's true for $q=1$ already we can assume that $q>1$. Then we can find a prime that divides $q$(**), so we can write $q = \tilde q p$ and we know that $\tilde q = \tilde y^2 \tilde z$ where is a product of distinct primes. Now $p$ is either one of those primes or it isn't. If it isn't we have that $\tilde zp$ is a product of distinct primes, that is $q = \tilde q p = \tilde y^2 (\tilde zp)$. Otherwise we have that $\tilde z/p$ is a product of distinct primes(***). So we have $q = (\tilde q p)^2 (\tilde z/p)$. So we have shown that it follows that it's true for $q$.
Now by induction it's true for all $q$.
(*) Can be proven by induction too:
What you use is to reformulate the induction step statement (using that $\psi \Rightarrow \phi$ implies that $\psi \Rightarrow \psi\land\phi$) as $(\forall j<n: P(x_j)) \Rightarrow (\forall j<n+1: P(x_j))$. Now you have a classic induction step and it follows that for all $n>1$ that $\forall j<n+1: P(x_j)$ and therefore especially that $P(x_n)$.
(**) Can be proven by induction too with the same extended induction principle (with the twist that we start at $2$ instead of $1$):
It's obviusly true for $2$. Now if it's true for all less than $q$ we have that since $q$ is composite that $q=ab$ where $b<q$ which in turn means that $b = cp$ for some prime $p$ so $q = (ac)p$ for some prime $p$. By extended induction it follows that all $q>1$ can be written $q=rp$ for some prime $p$.
(***) I guess that can be proven by induction too:)