A Möbius function product

Let $\mu$ denote the Möbius function. Prove that

$\prod_{n=1}^{\infty} \left ( 1 - x^n \right )^{-\mu(n)/n} = e^x \quad , \quad |x|<1$

Solution

Taking logarithms we have:

$\begin{align*} \log \prod_{n=1}^{\infty} \left ( 1 - x^n \right )^{-\mu(n)/n} &= -\sum_{n=1}^{\infty} \frac{\mu(n)}{n} \log \left ( 1 - x^n \right ) \\ &= \sum_{n=1}^{\infty} \frac{\mu(n)}{n} \sum_{m=1}^{\infty} \frac{x^{nm}}{m} \\ &= \sum_{n=1}^{\infty} \mu(n) \sum_{m=1}^{\infty} \frac{x^{nm}}{nm} \\ &= \sum_{m=1}^{\infty} \frac{x^m}{m} \sum_{d \mid m} \mu(d) \\ &= x + \sum_{m=2}^{\infty} \frac{x^m}{m} \sum_{d \mid m} \mu(d) \\ &= x \end{align*}$