When $0I am stuck after the following step.
$a^{n+1}<a^{x+1}b^{y}$ and $a^{x}b^{y+1}<b^{n+1}$.
Here $x+y+1=n+1$ but is it enough to prove the result by induction?
Just realised the statement could be rewritten as $a^{n}<a^{x}b^{n-x}<b^{n}$.
Couldn't prove it by induction but got an idea for an alternate proof from the comment of @kingW3 and the answer of @Arnaldo Nascimento.
Proof:
For $0<x$, $0<y$, and $0<n$.
$a^{y}<b^{y}\Rightarrow a^{y}a^{x}<b^{y}a^{x}\Rightarrow a^{n}=a^{x+y}<b^{y}a^{x}$ and $a^{x}<b^{x}\Rightarrow a^{x}b^{y}<b^{x}b^{y}\Rightarrow a^{x}b^{y}<b^{x+y}=b^{n}$. Therefore, $a^{n}<a^{x}b^{y}<b^{n}$.

$a^x=a^{n-y}=a^n.a^{-y}$ and $b^y=b^nb^{-x}$ so
$$a^xb^y=\frac{a^nb^n}{a^yb^x}<\frac{a^nb^n}{a^ya^x}=\frac{a^nb^n}{a^{x+y}}=b^n$$
$$a^xb^y=\frac{a^nb^n}{a^yb^x}>\frac{a^nb^n}{b^yb^x}=\frac{a^nb^n}{b^{x+y}}=a^n$$
and then
$$a^n<a^xb^y<b^n$$
