Given $$\prod_{k=0}^{n-1}(1+q^kt)=\sum_{k=0}^{n}q^{k(k-1)/2}\binom{n}{k}_{q}t^k$$ How do I show that as $n \rightarrow \infty$ $$\prod_{k=0}^{\infty}(1+q^kt)=\sum_{k=0}^{\infty}\frac{q^{k(k-1)/2} \ t^k}{\phi_k(q)}$$ where $\phi_k(q)=(1-q)(1-q^2)\cdots(1-q^k)$
It's give without proof on Wikipedia and in Macdonald's Symmetric Function and Hall Polynomials. Any kind of assistance will be appreciated
The $t^k$-coefficient is $$q^{k(k-1)/2}\binom nk_q=q^{k(k-1)/2}\frac{(1-q^n)(1-q^{n-1})\cdots(1-q^{n-k+1})}{(1-q)(1-q^2)\cdots (1-q^k)}.$$ As $n\to\infty$ the numerator tends to $1$, so this tends to $$\frac{q^{k(k-1)/2}}{(1-q)(1-q^2)\cdots (1-q^k)}.$$