Proving identity in infinite series I am trying to prove the equation
\begin{align*}
\sum_{n=0}^{\infty} a^n \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d} = \prod_{i=1}^d \frac{1}{1 - a b_i},
\end{align*}
where we have that $n = c_1 + c_2 + ... c_d$. This is what I have tried so far:
\begin{align*}
\sum_{n=0}^{\infty} a^{c_1 + c_2 + \ldots + c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d}
&= \sum_{n=0}^{\infty} a^{c_1} \cdot a^{c_2} \cdot \cdot \cdot a^{c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdot \cdot \cdot b_d^{c_d} \\
&= \sum_{c_1}^{\infty} a^{c_1} \cdots \sum_{c_d}^{\infty} a^{c_d} \cdot \sum_{c_1}^{} b_1^{c_1} \cdots \sum_{c_d} b_d^{c_d} \\
&= \sum_{c_1}^{\infty} (a b_1)^{c_1} \cdots \sum_{c_d}^{\infty} (ab_d)^{c_d}.
\end{align*}
My doubts in here are whether it is even possible to split up the sum in this manner and then simply recombining it in the last line, since the product of two sums is not necessarily the sum of the product.
From here I would use
\begin{align*}
\sum_{k}^{\infty} x^k = \frac{1}{1-x},
\end{align*}
such that it would become
\begin{align}
\frac{1}{1-a b_1} \times \cdots \times\frac{1}{1 - a b_d} = \prod_{i = 1}^{d} \frac{1}{1-a b_i}.
\end{align}
I would like to ask where my errors are and if there is a better way to prove this identity. Thank you for your time.

By induction on $d$.
$$\begin{aligned}
\frac{1}{1-ab_1}
\frac{1}{1-ab_2} &= \left(\sum_{n=0}^\infty (ab_1)^n\right)\left(\sum_{n=0}^\infty (ab_2)^n\right)\\
&=\sum_{n=0}^\infty \left(\sum_{l=0}^nb_1^lb_2^{n-l}\right) a^n\\
&=\sum_{n=0}^\infty \left(\sum_{c_1+c_2=n}b_1^{c_1} b_2^{c_2}\right) a^n
\end{aligned}$$
according to Cauchy product formula.
Now apply again Cauchy product formula to pass from $d$ to $d+1$:
$$\begin{aligned}
\prod_{i=1}^{d+1} \frac{1}{1 - a b_i} &=\left(\prod_{i=1}^{d} \frac{1}{1 - a b_i}\right)\frac{1}{1 - a b_{d+1}}\\
&= \left(\sum_{n=0}^\infty \left(\sum_{c_1+ \dots +c_d=n}b_1^{c_1} \dots b_d^{c_d}\right) a^n\right)\left(\sum_{n=0}^\infty (ab_{d+1})^n\right)\\
&= \sum_{n=0}^\infty \left(\sum_{l=0}^n \left(\sum_{c_1+ \dots +c_d=l}b_1^{c_1} \dots b_d^{c_d}\right)b_{d+1}^{n-l}\right)a^n\\
&=\sum_{n=0}^\infty \left(\sum_{c_1+\dots + c_{d+1}=n}b_1^{c_1} \dots b_{d+1}^{c_{d+1}}\right) a^n
\end{aligned}$$
