A combinatorial equality, maybe involving partition 
Prove the equality $$\prod_{n\geq 1} (1+tq^{2n-1})=\sum_{k\geq 0} \frac{t^k q^{k^2}}{(1-q^2)(1-q^4)...(1-q^{2k})}.$$

Attempt:
$t^k$ on the RHS indicates k terms of $tq^{2n-1}$ from the LHS are chosen for this summand. 
For $k=0$, only 1 are chosen in the LHS, so the corresponding summand is 1.
For $k=1$, the corresponding summand is $$tq+tq^3+tq^5+tq^7+....$$
For $k=2$, the corresponding summand is $$(tqtq^3+tqtq^5+tqtq^7+...)+(tq^3tq^5+tq^3tq^7+...)+(tq^5tq^7+...)+....$$
For $k=3$, the corresponding summand is $$tqtq^3tq^5+tqtq^3tq^7+tq^3tq^5tq^7+...$$
So we need to show that $$q^{n_1+n_2+...n_k}= \frac{q^{k^2}}{(1-q^2)(1-q^4)...(1-q^{2k})},$$ where $n_1,...,n_k$ are distinct odd numbers.
Question: Is the above argument a correct way? If so, how to finish the last step? I know the generating function of partition for odd summands is $\frac{1}{1-q}\frac{1}{1-q^3}\frac{1}{1-q^5}\cdot\cdot\cdot$, maybe it can start from here?
 A: I think your argument is correct. The following algebraic manipulation  should conclude the proof, although I do not know if this is the easiest way. 

Hint: Let $n_1<n_2<...<n_k$ be odd integers. Write $q^{n_1+n_2+...+n_k}$ as $$(q^{n_1})^k\cdot(q^{n_2-n_1})^{k-1}\cdot(q^{n_3-n_2})^{k-2}...(q^{n_k-n_{k-1}})$$ 

Now, $n_j$ are odd positive integers, so $n_1\in{}\{1,3,5,...\}$ and $n_{j+1}-n_j\in\{2,4,6,...\}$.
Thus
$$(q^{n_1})^k\in\{q^k,q^{3k},q^{5k},...\}$$
$$(q^{n_{j+1}-n_j})^{k-j}\in\{q^{2(k-j)},q^{4(k-j)},q^{6(k-j)},...\}$$
It follows that
$$\sum_{0<n_1<...<n_k \\\text{odd integers}} q^{n_1+n_2+..+n_k}=(q^k+q^{3k}+q^{5k}+..)(q^{2(k-1)}+q^{4(k-1)}+q^{6(k-1)}+..)(q^{2(k-2)}+q^{4(k-2)}+q^{6(k-2)}+..)...(q^2+q^4+q^6+...)$$ 
But the above sums are just geometric series.
$$\sum_{0<n_1<...<n_k \\\text{odd integers}} q^{n_1+n_2+..+n_k}=\frac{q^k}{1-q^{2k}}\cdot\frac{q^{2(k-1)}}{1-q^{2(k-1)}}\frac{q^{2(k-2)}}{1-q^{2(k-2)}}...\frac{q^2}{1-q^2}$$
$$=\frac{q^k\cdot q^{2(1+2+3+...+(k-1))}}{(1-q^2)(1-q^4)...(1-q^{2k})}$$
$$=\frac{q^k\cdot q^{k(k-1)}}{(1-q^2)(1-q^4)...(1-q^{2k})}$$
$$=\frac{q^{k^2}}{(1-q^2)(1-q^4)...(1-q^{2k})}$$
