Why every function can be written as a unique product of maxterms??
That's because each maxterm corresponds to one and only one tuple of values of the function's arguments. E.g., if a function f(x1, x2, x3) has three arguments, then the triple <x1, x2, x3> can have 8 values, from <0, 0, 0> to <1, 1, 1>. Each maxterm is true on one and only one of those 8 values.
Therefore, the sum-of-maxterms representation of a function f(x1, ..., xn) includes precisely those maxterms that correspond to values of <x1, ..., xn> where f(x1, ..., xn) = true.
Please post here if this explanation is unclear or if you need a more formal proof.
I am sorry, I thought you were talking about disjunctive normal form (DNF), not conjunctive normal form (CNF).
In CNF every maxterm corresponds to a line in the truth table where the function is false. First, for notation, suppose 0 means false and 1 means true. Next, if $\displaystyle x$ is a variable and $\displaystyle b=0$ or $\displaystyle b=1$, let $\displaystyle x^b$ denote $\displaystyle x$ when $\displaystyle b=1$ and $\displaystyle -x$ (the negation of $\displaystyle x$) when $\displaystyle b=0$.
Suppose $\displaystyle f(b_1,\dots,b_n)=0$ where $\displaystyle b_i=0$ or $\displaystyle b_i=1$. Then the maxterm $\displaystyle x_1^{-b_1}+\dots+x_n^{-b_n}$ in included into the CNF of $\displaystyle f$. E.g., if $\displaystyle f(0,1,0)=0$ then the maxterm $\displaystyle t_{010}=x_1+(-x_2)+x_3 $ is in the CNF. Indeed, $\displaystyle t_{010}=0$ if $\displaystyle x_1=0$, $\displaystyle x_2=1$ and $\displaystyle x_3=0$; on all other values of $\displaystyle x_1,x_2,x_3$ this maxterm is 1. Therefore, $\displaystyle t_{010}$ forces the whole conjunction to be false on 0,1,0 and is true (i.e., does not force anything) on other values.
From this it should be clear how to construct a CNF.
As for uniqueness, suppose $\displaystyle f=t_1\cdot\;\dots\;\cdot t_m=t_1'\cdot\;\dots\;\cdot t_k'=g$ where $\displaystyle t_i$ and $\displaystyle t_j'$ are maxterms. ($\displaystyle \cdot$ means conjunction). Then $\displaystyle f(b_1,\dots,b_n)=0$ implies thta there exist $\displaystyle 1\le i\le m $ such that $\displaystyle t_i(b_1,\dots,b_n)=0$. This means $\displaystyle t_i=x_1^{-b_1}+\dots+x_n^{-b_n}$ because every $\displaystyle x_j$ ($\displaystyle 1\le j\le n$) appears in $\displaystyle t_i$ and every member of disjunction must be false. Since the CNF's of $\displaystyle f$ and $\displaystyle g$ are the same, $\displaystyle t_i$ is one of $\displaystyle t_j'$ and therefore $\displaystyle g(b_1,\dots,b_n)$ is also 0. Similarly, $\displaystyle g(b_1,\dots,b_n)=0$ implies $\displaystyle f(b_1,\dots,b_n)=0$