A $q$-series identity: proof request Denote the $q$-expressions $[q]_n=(1-q)\cdots(1-q^n)$ for $n\geq1$ and $[q]_0:=1$. Also, $[q]_{\infty}=(1-q)(1-q^2)(1-q^3)\cdots$.

QUESTION. Is this identity true? It seems to be.
  $$\sum_{n=0}^{\infty}\left(1-\frac{[q]_{\infty}}{[q]_n}\right)
=\sum_{k=1}^{\infty}\frac{q^k}{1-q^k}.$$

 A: As darij grinberg says, the result is equivalent to the main result in the linked paper. On the right-hand side we have
$$\sum_{k=1}^\infty \frac{q^k}{1-q^k} = \sum_{k=1}^\infty (q^k + q^{2k} + \cdots ) = \sum_{n=1}^\infty \bigl( \sum_{d \mid n} 1 \bigr) q^n = \sum_{n=1}^\infty \tau(n) q^n $$
where $\tau(n)$ is the number of divisors of $n$. 
For the left-hand side, observe that $-[q]_\infty/[q]_{m-1} = -(1-q^m)(1-q^{m+1}) \cdots $ is the generating function for partitions with distinct parts not less than $m$, counting partitions with an even number of parts with sign $-1$. Therefore in $\sum_{m=1}^\infty (1- [q]_\infty/[q]_{m-1})$, a partition with smallest part $b(\lambda)$ is counted precisely $b(\lambda)$ times, once in each of the first $b(\lambda)$ summands, and always with the same sign. Therefore for $n \in \mathbb{N}$, the coefficient of $q^n$ in the left-hand side is
$$\sum_{\lambda \vdash n} (-1)^{\ell(\lambda)-1} b(\lambda). $$
By the main result of Bing, Fokkink and Fokkink, this is $\tau(n)$. Since the constant terms on either side are $0$, this proves the identity.
Remark. A well-known recurrence for the partition function is $np(n) = \sum_{m < n} \sigma(m)p(n-m)$, where $\sigma$ is the sum of divisors function.  MacMahon published two papers in the 1920s with many related identities, replacing $\sigma(n)$ with $\sigma_s(n) = \sum_{d \mid m} d^s$. But I cannot find this result in his work.
