Show that if $m$ and $n$ are integers and $a$ and $b$ are elements from a ring, then $(m \cdot a) ( n \cdot b) = (mn) \cdot (ab)$ 
Show that if $m$ and $n$ are integers and $a$ and $b$ are elements from a ring, then $(m \cdot a) ( n \cdot b) = (mn) \cdot (ab)$
Source:Gallian 7th ed Pg. 243, Q.15

Consider $P(n)$ to be the statement that  $\forall m \in M$ , $(m \cdot a) ( n \cdot b)=(mn) \cdot (ab)$ is true
Induction:

*

*Base case: $P(1) = (m \cdot a)( 1 \cdot b) = (m \cdot a)(b)$
Let $P(m)$ be the statement that $(m \cdot a)(b) = (m \cdot 1) (ab)$, then:
1.1. Base case : $(1 \cdot a)(b) =(ab)=  (1 \cdot 1)(ab)$
1.2. Proving $P(m+1)$ from $P(m)$: $ (m+1)a \cdot b= (m)a \cdot b + (1) a\cdot b= (m \cdot 1) ab + 1 a \cdot b= m(a \cdot b) + a \cdot b = (m+1) (a \cdot b)$


*Proving $n+1$th case from $nth$:
$P(n+1) = (  m \cdot a) \left( (n+1) \cdot b \right) = (m \cdot a) \left( n \cdot b + 1 \cdot b\right)= (m\cdot a)( n \cdot b) + ( m \cdot a) (b)= (mn) \cdot (a b) + (m) \cdot (ab)= (mn+m) \cdot (ab)=\left( (m+1)n\right) \cdot (ab)$
Remark: I have used the result in the sub induction to simplify the last equality in the above.
I am not sure how to extend the above proof to $m,n \in \mathbb{Z}$.

Could I get some one check my work so far, and, explain how I could finish it?

Perhaps show the simpler statements
$$\tag1(n\cdot a)\cdot b=n\cdot(a\cdot b)$$
and
$$\tag2a\cdot(n\cdot b)=n\cdot(a\cdot b)$$
and
$$\tag3m\cdot(n\cdot a)=(mn)\cdot a$$
(which you may already have done).
Then use these to see
$$(m\cdot a)\cdot(n\cdot b)
\stackrel{(1)}=m\cdot(a\cdot(n\cdot b))
\stackrel{(2)}=m\cdot(n\cdot(a\cdot b))
\stackrel{(3)}=(mn)\cdot(a\cdot b)$$
