Let $G$ be a group. Show that $\forall a, b, c \in G$, the elements $abc, bca, cab$ have the same order.
In general, conjugate elements have the same order:
$o(ghg^{-1}) = o(h)$
because $ x \mapsto gxg^{-1}$ is an automorphism.
Now note that $bca = a^{-1} (abc) a$ and $cab = c (abc) c^{-1}$ are conjugates of $abc$.
Notice that you just need to show that $ o(ab) = o(ba) $. Let $ n=o(ab) $. You have
$$ (ba)^{n+1} = b(ab)^{n}a = ba $$
Thus, by multiplying by $ (ba)^{-1} $, you get $ o(ba) \leq o(ab) $. Symmetry then gives the desired equality.
In your proof, you only showed that $ o(ba) \leq o(ab)$.
Two things you should consider:
Your first step only works if $n \ge 2$. You should handle the case $n = 1$ separately.
You've shown that if $(abc)^n = 1$ then $(bca)^n = e$ and $(cab)^n = e$. What this shows is that $|abc| \ge |bca|$ and $|abc| \ge |cab|$ because, at least in principle, $(bca)^m$ might equal $e$ for some $m < n$.