If $a=b^c$, $b=c^a$, and $c=a^b$, prove that If $$a=b^c, \quad b=c^a, \quad \text{and} \quad c=a^b$$ prove that $$abc=1.$$
My Attempt;
Given,
$$a=b^c$$
$$b=c^a$$
$$c=a^b$$
Now,
$$
\begin{align}
\text{L.H.S.} & =abc \\
& = b^c\cdot b\cdot a^b \\
& =b^c\cdot b\cdot b^{bc} \\
& =b^{c+1+bc} \\
& \,\,\,\vdots
\end{align}
$$
I got struck from here.
Please help to complete.

In $a=b^c$, substitute $b=c^a$, and you get $a=(c^a)^c=c^{ac}$. Next substitute $c=a^b$ and get $a=(a^b)^{ac}=a^{abc}$. So $a^{1}=a^{abc}$. Either $abc=1$ or $a=1$.
If $a=1$ then $c=a^b=1$ and $b=c^a=1$, so $abc=1$ again.
