If you know that whenever $a\lt b$ and $c\gt 0$ then $ac\lt bc$, then for any $x$ with $0\lt x\lt 1$, multiplying through by $x$ gives
\begin{equation*}
0 \lt x^2 \lt x \lt 1.
\end{equation*}
(The last inequality is "inherited" from the fact that you already know that $x\lt 1$). Repeat to get
\begin{equation*}
0\lt x^3 \lt x^2 \lt x \lt 1.
\end{equation*}
And so on; you get $0\lt x^n \lt x^{n-1}\lt\cdots \lt x \lt 1$.
Working the other way, if $1\lt y$, then multiplying through by $y$ you get
\begin{equation*}
1\lt y\lt y^2
\end{equation*}
(with the first inequality "inherited from your original assumption) and repeating this process leads to
\begin{equation*}
1\lt y\lt y^2 \lt\cdots \lt y^n
\end{equation*}
So: if $0\lt z\lt 1$, then $0\lt z^n\lt 1$. If $1\lt z$, then $1\lt z^n$. And of course, if $z=1$ then $z^n = 1$.
So, start with $0\lt a\lt 1$. Then $\sqrt[n]{a}$ cannot be greater than $1$ (then $a = \left(\sqrt[n]{a}\right)^n \gt 1$ by the above), so $\sqrt[n]{a}\lt 1$.
Now replace $x$ with $\sqrt[n]{a}$ to get the desired inequality.