Below are equivalent definitions of $\rm\ q\,$ squarefree. Yours is $(5)$.
Theorem $\ $ Let $\rm\ 0 \ne q\in \mathbb Z\:.\ \ $ The following are equivalent.
$(1)\rm\quad\ \ \ \, n^2\,|\ q\ \ \Rightarrow\ \ n\ |\ 1\qquad\ $ for all $\rm\:\ n\in \mathbb Z $
$(2)\rm\quad\ \ \ \, n^2\, |\, qm^2 \!\Rightarrow n\ |\ m\qquad\! $ for all $\rm\: \ n,m\in \mathbb Z$
$(3)\rm\qquad\ q\ |\ n^2\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z $
$(4)\rm\qquad\ q\ |\ n^k\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb Z,\ k\in \mathbb N $
$(5)\rm\quad\:\ \: q^q\ |\ n^n\ \Rightarrow\ q\ |\ n\qquad\ $ for all $\rm\:\ n\in \mathbb N,\ $ for $\rm\ q > 0 $
Proof $\ \: (1\Rightarrow 2)\rm\:\ \ $ Canceling $\rm\:(n,m)^2\:$ from LHS of $(2)\:$ we may assume w.l.o.g. that $\rm\:(n,m)\:=\:1.\ $ By $ $ Euclid's Lemma $\rm\: n^2\, |\, qm^2\: \Rightarrow\ n^2\: |\: q\ \Rightarrow\ n\:|\:1\ \Rightarrow\ n\:|\:m$
$(2\Rightarrow 3)\rm\quad q\ |\ n^2\ \Rightarrow\ q^2\ |\ qn^2\ \Rightarrow\ q\ |\ n $
$(3\Rightarrow 4)\rm\quad k \ge 2\ \Rightarrow\ k \le 2\:(k-1)\ $ so $\rm\:\ q\ |\ n^k\ |\ (n^{k-1})^2\ \Rightarrow\ q\ |\ n^{k-1}\:\ldots\:\Rightarrow\ q\ |\ n$
$(4\Rightarrow 5)\rm\quad q\ |\ q^q\ |\ n^n\ \Rightarrow\ q\ |\ n $
$(5\Rightarrow 1)\:$ via $\:\lnot\: 1\Rightarrow\lnot\: 5.\ $ By $\rm\:\lnot 1,\,\ q\: =\: ab^2,\:\ b\nmid 1.\:$ Put $\rm\ n = abc\:$ for $\rm\:c\:$ as below.
$\rm\qquad\ \ \ q\ |\ (ab)^2\ \Rightarrow\ q^{\:q}\ |\ (ab)^{2\:q}\ |\ (abc)^{abc}\! = n^n\quad\ \ for\ all \:\ c\:\ with\ \ abc > 2\:q $
Since $\rm\ b\nmid 1\ \Rightarrow\ q\nmid ab,\:$ we may choose $\rm\:c\:$ so that also $\rm\ q\nmid abc,\ $ e.g. $\rm\:\ c\equiv 1\,\ (mod\ q)$