Proof of transfer function factorization $\frac{b_0}{a_0} \frac{\prod_{k=1}^M (1-c_kz^{-1})}{\prod_{k=1}^N(1-d_kz^{-1})}$
It's a mere application of the fundamental theorem of algebra of polynomials in 1D, which states that any polynomial $p(x)$ of $x$ of degree $n$, can be written (expressed) in either increasing (or decreasing) powers of $x$ as in $$p(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n = \sum_{k=0}^{n}{a_kx^{n-k}}$$  or as a product of its $n$ roots in the factors of $(x-r_k)$ as in $$p(x) = K(x-r_1)(x-r_2)...(x-r_n)=K\prod_{k=1}^{n}(x-r_n)$$ where each such factor represents a root of the polynomial $p(x)$. 

The $n^{th}$ degree poynomial is assumed to have $n$ roots, either distinct or repeating or in complex and/or complex conjugate forms, depending on the coefficients $a_k$.

Applying this to the $Z$-Transform of the form discussed:
$$X(z)=\sum_{n=0}^{M}{a_nz^{-n}}=a_0+a_1z^{-1}+...+a_Mz^{-M}$$ this can be seen either as a polynomial in the variable $z^{-1}$, or after factoring out $a_0z^{-M}$ ($a_0$ being nonzero) it will be a polynomials in $z$ as given $$X(z)=a_0z^{-M}\left(z^{M}+\frac{a_1}{a_0}z^{M-1}+...+\frac{a_M}{a_0}\right)$$ And the paranthesis can be written in product form as: 
$$X(z)= a_0z^{-M}\prod_{k=1}^{M}{(z-r_k)}$$
You can distribute the factor $z^{-M}$ into each product form as in: $$z^{-1}(z-p_m) = (1-p_mz^{-1}) $$ which yields the final required form as: 
$$X(z) = a_0 \prod_{k=1}^{M}(1 - r_k z^{-1})$$ where each $r_k$ is a root.

When $X(z)$ is expressed as a rational traform as in $$X(z)= \frac{P(z)}{Q(z)} = \frac{\sum_{k=0}^{M}{b_kz^{-k}}}{\sum_{k=0}^{N}{a_kz^{-k}}} $$ apply this factorization to both of numerator and denominator to get the final form:$$X(z)= \frac{\sum_{k=0}^{M}{b_kz^{-k}}}{\sum_{k=0}^{N}{a_kz^{-k}}}=\frac{b_0}{a_0} \frac{\prod_{k=1}^{M}{(1-c_kz^{-1})}}{\prod_{k=1}^{N}{(1-d_kz^{-1})}}$$