Proof that $\log_a b \cdot \log_b a = 1$ 
Prove that $\log_a b \cdot \log_b a = 1$

I could be totally off here but feel that I have at least a clue.
My proof is: 
Suppose that $a = b$, then $a^{1} = b$ and $b^{1} = a$ and we are done.
Suppose now that $a \neq b$. We wish to show that: $\log_b a = \frac{1}{\log_a b}$.
\begin{align*}
1 &= \frac{1}{\log_a b} \cdot \frac{1}{\log_b a} \cdot \left(\log_a{b} \cdot \log_b a \right)    \\
   &=  \left(\log_a{b} \cdot \frac{1}{\log_a b} \right) \cdot  \left(\log_b a \cdot \frac{1}{\log_b a}\right)  \\
   &=  \left(\log_b a \cdot \frac{1}{\log_b a}\right) \cdot 1    \\
   &=1 \cdot 1    \\
  &= 1
\end{align*} 

Method I
Let $\log_ab=x$ and $\log_ba=y$. Then we have
$$a^x=b \quad\text{and}\quad b^y=a$$
So,
\begin{align}
(b^y)^x&=b\\
b^{xy}&=b\\
xy&=1
\end{align}

Method II
Let $\log_ab=x$. Then we have
\begin{align}
a^x&=b\\
\log_b(a^x)&=\log_bb\\
x\log_ba&=1
\end{align}
