A determinant

Let $(a, b)$ denote the great common divisor of the numbers $a, b$. Prove that:

$$\begin{vmatrix}
\left ( 1,1 \right ) &\left ( 1,2 \right )  &\cdots  &\left ( 1,n \right ) \\
 \left ( 2,1 \right )&\left ( 2,2 \right )  & \cdots &\left ( 2,n \right ) \\
 \vdots & \vdots  & \ddots & \vdots \\
 \left ( n,1 \right )& \left ( n, 2 \right ) &\cdots  & \left ( n,n \right )
\end{vmatrix} = \prod_{k=1}^{n}\phi(k)$$

where $\phi$ is Euler's phi function.

Solution

We set $f(n)=n$ . Then:

$$f(n)=n= \sum_{d \mid n} g(d)\overset{\text{Mobius  Transformation}}{=\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! \Rightarrow }\;  g(n)= \sum_{d \mid n} \mu (d)f \left ( \frac{n}{d} \right )= \\
\sum_{d \mid n} f(d) \mu \left ( \frac{d}{n} \right ) = \sum_{d \mid n}d \mu \left ( \frac{d}{n} \right )$$

Hence $f$ is totally multiplicative. However, it holds that:

$$\left | \left ( i,j \right ) \right |= \prod_{j=1}^{n}j \prod_{p \mid j} \left ( 1- \frac{1}{p} \right )= \prod_{k=1}^{n}\phi(k)$$