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

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.
