I need to prove that $(a^n)(b^n) = (ab)^n,$ where $a,b\in\mathbb N$. Proof by induction on $n \in \mathbb Z, n\geq 0$.

Question

I need to prove that $(a^n)(b^n) = (ab)^n,$ where $a,b\in\mathbb N$.

Since $a^0=1$, is given in the question, I assumed my base step to be when $(n=0)$.
$(a^0)(b^0)= (1)(1) =1.$
and now I'm stuck.
I also tried
$(a^{n+1})(b^{n+1})=(ab)^{n+1}$
as my induction.
In the question, $a^{n+1}$ is given as $(a^n)(a)$
I got to
$(a^{n+1})(b^{n+1})=(a^na)(b^nb)$
and I'm stuck again.
Help?

Answer 1

I will assume you are allowed to use other laws such as the commutative law of multiplication.

As you mentioned before $$a^nb^n=(ab)^n$$ for $n=0$ Now assume that it holds for $n=k$, then for $n=k+1$ we have $$a^{k+1}b^{k+1}=a^kab^kb=a^kb^kab=(ab)^kab$$ Now recall that all exponent raised to the first power is itself, therefore $ab=(ab)^1$

$$(ab)^k(ab)^1$$

Multiplication of exponents witht he same base

$$(ab)^k(ab)^1=\boxed{(ab)^{k+1}}$$

Answer 2

multiplication is commutative and associative.

$(a^{n+1})(b^{n+1}) = (a^na)(b^nb) = a^n\cdot a\cdot b^n \cdot b= a^n\cdot b^n\cdot a\cdot b=$

$(a^nb^n)(ab) = (ab)^n (ab)= (ab)^{n+1}$.