Proof: By Induction

For $n,m,p\in\mathbb N$ we will first prove the "right-distributivity law" $(n+m)\cdot p=(n\cdot p)+(m\cdot p)$. The validity of the "left-distributivity law" $p\cdot(n+m)=(p\cdot n)+(p\cdot m)$ then follows from the commutativity of the multiplication of natural numbers.
Let $n,m,p\in\mathbb N$ be arbitrary natural numbers.
Considering the introduced binary operations on natural numbers addition "$ + $" and multiplication "$\cdot$", we will prove the right-distributivity law $(n+m)\cdot p=(n\cdot p)+(m\cdot p)$ by induction over $p\in\mathbb N$.

Let the rule $(n+m)\cdot p=(n\cdot p + m\cdot p)$ be proven for all $p\ge 0$.
According to the definition of multiplication of natural numbers we have $$(n+m)\cdot p^+:=(n+m)\cdot p + (n+m).\quad\quad( * )$$
On the other hand, it follows from the commutativity of the addition of natural numbers that $$\begin{array}{ccl} (n\cdot p^+ + m\cdot p^+)&=&n\cdot p + n + m\cdot p + m\\ &=&n\cdot p + m\cdot p + n + m,\end{array}$$ which equals $ ( * ) $, by hypothesis.
It follows that $(n+m)\cdot p^+=(n\cdot p^+ + m\cdot p^+).$
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Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013