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.
 A: 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.
A: $$abc = a^bb^cc^a$$
$$\implies \log(abc) = \log(a^bb^cc^a)$$
$$\implies \log(abc) = \log b^c + \log c^a + \log a^b$$
$$\implies \log(abc) = c\log b + a\log c + b\log a$$
$$\implies \log(abc) = c\log c^a + a\log a^b + b\log b^c$$
$$\implies \log(abc) = ac\log c + ab\log a + bc\log b$$
$$\implies \log(abc) = ac\log a^b + ab\log b^c + bc\log c^a$$
$$\implies \log(abc) = abc\log a + abc\log b + abc\log c$$
$$\implies \log(abc) = abc(\log a + \log b + \log c)$$
$$\implies \log(abc) = abc\log(abc)$$
$$\implies 1 = abc$$  
Not the shortest proof, but I consider it by far the most beautiful I could think of due to the symmetry in the logarithm steps!
A: Another answer that involves logarithms.
if $a=b^c$, $b=c^a$, $c=a^b$ then we may write that
$c=\log_{b}(a)$,$a=\log_{c}(b)$ and $b=\log_{a}(c)$
Hence we have that $abc=\log_{b}(a)\log_{c}(b)\log_{a}(c)$. Then by using change of base rule we get
$abc=\frac{\log_{e}(a)}{\log_{e}(b)}\frac{\log_{e}(b)}{\log_{e}(c)}\frac{\log_{e}(c)}{\log_{e}(a)}$
Notice that all the logarithms on the right hand side cancel out and we get the desired result. 
A: $$
a^1 = a = b^c = (c^a)^c = c^{ac} = (a^b)^{ac} = a^{bac},
$$
so
$$
a^1 = a^{bca}.
$$
Now take base-$a$ logarithms of both sides.
A: Actually, I note that $a=b=c=-1$ works just as well, which contradicts everyone's statements that $abc=1$.
