How to demonstrate this rule of multiplication? Let $S(n)$ be the sum of the digits of $n \ge 0$. Let $R(n)$ be the reduced form of $n$, that is, apply $S(n)$ and if the result is greater or equal than $10$, then apply $S(n)$ again and repeat until the result is less than $10.$ For example, $R(845) = R(17) = 8$.
I want to demonstrate, being $n,m \ge 0$, that
$$R(n*m) = R(R(n)*R(m))$$
I thought first of showing that $R(n+m) = R(R(n)+R(m))$, and if it is true, extend it to more terms, finally concluding that:
(the number of terms is $m$)
$$R(n*m) = R(n+n+..+n) = R(R(n)+R(n)+...+R(n)) = R(R(n)*m)$$
and then, by the same reasoning:
(the number of terms is $R(n)$)
$$R(R(n)*m) = R(m+m+..+m) = R(R(m)+R(m)+...+R(m)) = R(R(m)*R(n))$$
But I can't prove $R(n+m) = R(R(n)+R(m))$, and neither I know how to extend it to more terms assuming it is true.
I took this into account: $R(S(n)) = S(R(n)) = R(n) = R(R(n))$.
 A: You're making the problem much too hard.  All you need to do is consider 100 cases, because $R(n)$ and $R(m)$ are each in the range $0..9$.
A: $R(n)$ is the unique $x\in \{0,1,...,8\}$ such that $n-x$ is a multiple of $9.$ That is, $n=R(n)+9k$ for some $k\in \Bbb Z.$ In particular $R(n)=0$ iff $9$ divides $n.$ 
For any natural numbers $A,B$ we have $R(A)=R(B)$ iff $9$ divides $(A-B).$
Proof: Let $A=R(A)+9k$ and $B=R(B)+9l$ with integers $k,l.$ 
(i). Suppose $R(A)=R(B).$ Then $A-B=(R(A)-R(B))+9(k-l)=9(k-l)$ is divisible by $9.$
(ii). Suppose $9$ divides $A-B.$ Then $9$ divides $(R(A)-R(B)+9(k-l)$ so $9$ divides $R(A)-R(B).$  But $R(A)$ and $R(B)$ belong to $\{0,1,...,8\}$ so $-8\leq R(A)-R(B)\leq 8.$  And the only multiple of $9$ in the range $-8,-7,..., 7,8$  is the number $0,$ so $R(A)-R(B)$ (which is a multiple of $9$) must equal $0.$ (End of proof.)
THEREFORE: If $n=R(n)+9k$ and $n'=R(n)+9k'$ with integers $k,k'$  then   $$nn'-R(n)R(n')=(R(n)+9k)(R(n')+9k')-R(n)R(n')=9(kR(n')+k'R(n)+kk')$$ is divisible by $9.$ ......  So $R(nn')=R(R(n)R(n')).$
