$q$-series and Stirling of the 1st kind

Question

Denote the (unsigned) Stirling numbers of the $1^{st}$-kind by ${n \brack k}$ and define $$\mathbf{F}_a(q)=\sum_{m\geq1}\frac{q^{am}}{(1-q^m)^{2a}} \qquad \text{and} \qquad \mathbf{G}_b(q)=\sum_{m\geq1}\frac{m^bq^m}{1-q^m}.$$

I have encountered the following property.

QUESTION. Is this true? Or, can you provide a reference to it. $$\mathbf{F}_a(q)=\frac1{(2a-1)!}\sum_{k=0}^{a-1}(-1)^k\left(\sum_{j=-k}^k(-1)^j {a \brack a-k+j} \cdot {a\brack a-k-j}\right)\mathbf{G}_{2a-1-2k}(q).$$

Example. When $a=2$, the claim reads as $$\sum_{m\geq1}\frac{q^{2m}}{(1-q^m)^4} =\frac16\sum_{m\geq1}\frac{m^3q^m}{1-q^m}-\frac16\sum_{m\geq1}\frac{mq^m}{1-q^m}.$$ Observe that the right-hand side enumerates excess in the sum of powers of divisors.

Postscript. I have found a better format.

QUESTION. Is this true? Or, can you provide a reference to it. $$\mathbf{F}_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$ where we adopt an umbral notation $\mathbf{G}^j$ to stand for $\mathbf{G}_j$ and multiply accordingly.

Answer 1

QUESTION. Is this true? Or, can you provide a reference to it. $$\mathbf{F}_a(q)=\frac1{(2a-1)!}\mathbf{G}(\mathbf{G}^2-1^2)(\mathbf{G}^2-2^2)(\mathbf{G}^2-3^2)\cdots(\mathbf{G}^2-(a-1)^2);$$ where we adopt an umbral notation $\mathbf{G}^j$ to stand for $\mathbf{G}_j$ and multiply accordingly.

Expanding the differences of two squares this is $$\mathbf{F}_a(q) \stackrel?= \frac1{(2a-1)!}(\mathbf{G}+a-1)^{\underline{2a-1}}$$ and expanding the umbral definition of $\mathbf{G}$ this becomes $$\mathbf{F}_a(q) \stackrel?= \frac1{(2a-1)!} \sum_{m \geq 1}\frac{(m+a-1)^{\underline{2a-1}} q^m}{1-q^m} = \sum_{m \geq a} \binom{m+a-1}{2a-1} \frac{q^m}{1-q^m}$$ where the range of $m$ in the final sum has been tweaked to the support of the binomial.

It's standard that $\frac{1}{(1-x)^k} = \sum_{j \ge 0} \binom{j+k-1}{k-1} x^j$, so

$$\mathbf{F}_a(q) = \sum_{m \geq 1}\frac{q^{am}}{(1-q^m)^{2a}} = \sum_{m \geq 1} \sum_{j \ge 0} \binom{j+2a-1}{2a-1} q^{(j+a)m} $$

Subst $k = j+a$:

$$\mathbf{F}_a(q) = \sum_{m \geq 1} \sum_{k \ge a} \binom{k+a-1}{2a-1} q^{km} = \sum_{k \ge a} \binom{k+a-1}{2a-1} \sum_{m \geq 1} q^{km} = \sum_{k \ge a} \binom{k+a-1}{2a-1} \frac{q^k}{1 - q^k} \blacksquare$$