A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0.3cm} \log_{c}{a} = \frac{\log_{a}{a}}{\log_{a}{c}}\\~\\
\rightarrow \log_{b}{a} = \frac{\log_{b}{b}}{\log_{a}{b}}, \hspace{0.3cm} \log_{c}{b} = \frac{\log_{c}{c}}{\log_{b}{c}}, \hspace{0.3cm} \log_{a}{c} = \frac{\log_{a}{a}}{\log_{c}{a}}\\~\\
\rightarrow \large\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\rightarrow\\~\\
\rightarrow\frac{\log_{b}{b}}{\log_{a}{b}} \times \frac{\log_{c}{c}}{\log_{b}{c}} \times \frac{\log_{a}{a}}{\log_{c}{a}} \overset? = 1\\
\rightarrow \frac{1}{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}} \overset? = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} \overset? = \log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\\
\rightarrow \frac{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}}{\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}} = 1 \\
\rightarrow \frac{\log_{a}{b}}{\log_{b}{a}} \times \frac{\log_{b}{c}}{\log_{c}{b}} \times \frac{\log_{c}{a}}{\log_{a}{c}} = 1\\~\\
\small \text{------ Using the ‘change of base rule' again ------} \\~\\
\Large \therefore \hspace{0.2cm}\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1
 $
Any other methods?
Thanks in advance.
 A: Well it is much simpler. By definition 
$$\log_b(a) = \frac{\ln(a)}{\ln(b)}, \quad \log_c(b) = \frac{\ln(b)}{\ln(c)}, \quad log_a(c) = \frac{\ln(c)}{\ln(a)}$$
where $\ln$ is the logarithm in base $e$.
Your identity directly follows.
A: You may show it directly: let
$$x=\log_a(c)\log_c(b)\log_b(a),$$
hence
\begin{align}a^{\color{red}{x}}&=a^{\log_a(c)\log_c(b)\log_b(a)}\\
&=\left(\left(a^{\log_a(c)}\right)^{\log_c(b)}\right)^{\log_b(a)}\\
&=(c^{\log_c(b)})^{\log_b(a)}\\
&=b^{\log_b(a)}\\
&=a\\
&=a^{\color{red}{1}}.
\end{align}
A: You can generalize this result with logarithm rule: 
$\log_{b} a = \dfrac{\log_{e}a}{\log_{e}b} = \dfrac{\ln a}{\ln b}$. That is, the identity is valid for all $a_i$ positives reals numbers and differents of $1$:
$$
\log_{a_2}a_1 \cdot \log_{a_3}a_2 \cdot \log_{a_4}a_3 \cdots \log_{a_n}a_{n-1}\cdot \log_{a_1}a_n = 1.
$$
Firt all, we have for $n=2$, implies $\log_{a_2}a_1 \cdot \log_{a_1}{a_2} = 
\dfrac{\ln a_1}{\ln a_2}\cdot \dfrac{\ln a_2}{\ln a_1} = 1.$
Your case is for $n=3$, that is
$$
\log_{a_2}a_1 \cdot \log_{a_3}{a_2} \cdot \log_{a_1}{a_3} = 
\dfrac{\ln a_1}{\ln a_2}\cdot 
\dfrac{\ln a_2}{\ln a_3} \cdot 
\dfrac{\ln a_3}{\ln a_1} = 1.
$$ 
In general, suposse that the identity is valid for some natural number $n$, and we need to prove the valid for $n+1$. Then
$$
\log_{a_2}a_1 \cdot \log_{a_3}a_2 \cdot 
\log_{a_4}a_3 \cdots \log_{a_n}a_{n-1}\cdot \log_{a_{n+1}}a_n \cdot \log_{a_1}{a_{n+1}} =\\ 
\dfrac{\ln a_1}{\ln a_2}\cdot 
\dfrac{\ln a_2}{\ln a_3} \cdot 
\dfrac{\ln a_3}{\ln a_4} \cdots
\dfrac{\ln{a_{n-1}}}{\ln{a_n}} \cdot
\dfrac{\ln{a_{n}}}{\ln{a_{n+1}}} \cdot
\dfrac{\ln{a_{n+1}}}{\ln{a_1}} =1.
$$
A: $$\log_{b}{a}\log_{c}{b}\log_{a}{c}=e^{\ln\left(\log_{b}{a}\log_{c}{b}\log_{a}{c}\right)}=$$
$$e^{\ln\ln{a}-\ln\ln{b}+\ln\ln{b}-\ln\ln{c}+\ln\ln{c}-\ln\ln{a}}=e^0=1$$
