Proof that Category Product is a Category let A and B be categories, and let's define category product as:

$A \times B \;\;$ where:

$(1)$ the objects are pairs $(a, b)$ such that $a \in A$ and $b \in B$.

$(2)$ the arrow work like $(f,g) : (A, B) \rightarrow (C, D)$ where:


$f \in A, \;\;f:A \rightarrow C$    
$g \in B, \;\;f:B \rightarrow D$


$(3)$ the identity is defined as $\;1_{(A,B)} = (1_{A}, 1_{B})$

$(4)$ the composition is defined as $(f, g)\circ (f',g')$ = $(f\circ f',\;g\circ g')$



I think I should prove it like this:

1 - Prove that the identity is identity for composition operation

2 - Prove that the composition is associative.



[1] let $(f,g) = fg $

$fg = fg \circ 1_{(A,B)}$

so

$fg( (a,b) ) = fg \times 1_{(A,B)}(a,b)$

$fg( (a,b) ) = fg ((a,b))$



my questions are:


do I have to prove what I said I should?
is the identity proof correct?
how can I prove associativity over composition?
Why would you set $(f,g)=fg$? That looks like you're confusing a morphism $(f,g)$ in the product with the composition $fg$ of two morphisms in the same category. Due to this confusion, none of the rest of your argument makes sense. You need to prove, for any $(f,g):(A,B)\to (A',B'),$ that $(f,g)=(f,g)\circ 1_{A,B}=1_{A',B'}\circ (f,g)$.

The proof should look something like this:
$(f,g)\circ 1_{A,B}=(f,g)\circ (1_A,1_B)=(f\circ 1_A,g\circ 1_B)=(f,g)$, and similarly for sticking the identity on the other side. Then you should do the same kind of thing for associativity.
Associativity is a statement about three morphisms, so it doesn't look like you've checked that here.  The statements are true and the definition is standard; see https://ncatlab.org/nlab/show/product+category