If $A$ and $B$ are square matrices of the same order such that $AB = BA$,then prove by induction that $AB^{n} = B^{n}A$. Further,prove that $(AB)^{n} = A^{n}B^{n}$ for all $n \in N$.

(N/A) Given that $A$ and $B$ are square matrices of the same order such that $AB = BA$.
Part $1$: To prove $P(n): AB^{n} = B^{n}A$ for all $n \in N$ by induction.
For $n = 1$,$AB^{1} = B^{1}A$,which is true as $AB = BA$ is given.
Assume the result is true for $n = k$,i.e.,$AB^{k} = B^{k}A$ $(1)$.
For $n = k + 1$,we have $AB^{k+1} = (AB^{k})B = (B^{k}A)B = B^{k}(AB) = B^{k}(BA) = (B^{k}B)A = B^{k+1}A$.
Thus,by the principle of mathematical induction,$AB^{n} = B^{n}A$ for all $n \in N$.
Part $2$: To prove $Q(n): (AB)^{n} = A^{n}B^{n}$ for all $n \in N$ by induction.
For $n = 1$,$(AB)^{1} = A^{1}B^{1} = AB$,which is true.
Assume the result is true for $n = k$,i.e.,$(AB)^{k} = A^{k}B^{k}$ $(2)$.
For $n = k + 1$,we have $(AB)^{k+1} = (AB)^{k}(AB) = (A^{k}B^{k})(AB) = A^{k}(B^{k}A)B$.
Using the result from Part $1$,$B^{k}A = AB^{k}$,so $(AB)^{k+1} = A^{k}(AB^{k})B = (A^{k}A)(B^{k}B) = A^{k+1}B^{k+1}$.
Thus,by the principle of mathematical induction,$(AB)^{n} = A^{n}B^{n}$ for all $n \in N$.