Let $m,n$ be natural numbers. Then $n\times m = m\times n$.
MY ATTEMPT (EDIT)
Lemma 1
We shall need first the following result: $m\times 0 = 0$.
Let us prove it by induction on $m$. Indeed, one has that $0\times 0 = 0$, by the definition of multiplication by $0$ on the left. Let us assume the proposition holds for $m$, that is to say, $m\times 0 = 0$, and we shall prove it to $m\texttt{+}\texttt{+}$. Indeed, one has \begin{align*} (m\texttt{+}\texttt{+})\times 0 = (m\times 0) + 0 = 0 + 0 = 0 \end{align*}
And we are done.
Lemma 2
We shall prove that $m\times(n\texttt{+}\texttt{+}) = m\times n + m$ by induction on $m$.
To start with, notice that $0\times(n\texttt{+}\texttt{+}) = 0 = 0\times n + 0$, and the base case is done. Let us assume that $m\times(n\texttt{+}\texttt{+}) = m\times n + m$, and prove it holds that $(m\texttt{+}\texttt{+})\times(n\texttt{+}\texttt{+}) = (m\texttt{+}\texttt{+})\times n + m\texttt{+}\texttt{+}$: \begin{align*} (m\texttt{+}\texttt{+})\times(n\texttt{+}\texttt{+}) & = m\times(n\texttt{+}\texttt{+}) + n\texttt{+}\texttt{+} = m\times n + m + n \texttt{+}\texttt{+}\\\\ & = (m\times n + n\texttt{+}\texttt{+}) + m = (m\times n + n)\texttt{+}\texttt{+} + m\\\\ & = ((m\texttt{+}\texttt{+})\times n)\texttt{+}\texttt{+} + m = ((m\texttt{+}\texttt{+})\times n + m)\texttt{+}\texttt{+}\\\\ & = (m\texttt{+}\texttt{+})\times n + m\texttt{+}\texttt{+} \end{align*} And we are done.
Proposition
Based on the previous result, we shall prove the proposed statement by induction on $n$. According to lemma 1, one has that $0\times m = m\times 0 = 0$. Let us assume that $n\times m = m\times n$ and let us prove it for $n\texttt{+}\texttt{+}$. Second lemma 2, we have that \begin{align*} (n\texttt{+}\texttt{+})\times m = n\times m + m = m\times n + m = m\times(n\texttt{+}\texttt{+}) \end{align*}
And we are done.
Any comments or contributions on the solution?
$(n++)\times m=(n\times m)+ m = (m\times n)+m=m+(m\times n)=m\times (n++)$