$a = d$ implies $a^b = d^b$

Question

Prove that $a = d$ implies $a^b = d^b$, where $a, d$ are arbitrary nonnegative integers and $b$ is any positive integer.

If I could use division I think it could be something like that: $a^b / d^b = a ^{b-b} = a^0 = 1$ (assuming $a = d$), but I'm trying to figure out how to prove this using only multiplication and addition properties (natural numbers).

Here's my idea:

If a = b then there exists an integer $k$ such that $x = ak$ and $x = bk$, that also implies $(a^b)k = (d^b) k$ and $a^b = d^b$ by using the cancellation law of multiplication.

Answer 1

You're overthinking it. We proceed by induction on $b$. The case $b=1$ is trivial, so assume it works when $b=k$. Then $a^{k+1}=a^ka=d^kd=d^{k+1}$ completes the inductive step.

Answer 2

There must be something I don't understand. Are you sure you have asked the question you intended?

The equals sign in the assumption $a=d$ means that "$a$" and "$d$" are essentially just different names for the same number. So you can substitute one for the other in any formula. There is nothing to prove.