Use mathematical induction to prove the property for all positive integers n.
For the equation: [a^n]^5=a^(5n)
This makes no sense to me at all.
For the equation: [a^n]^5=a^(5n)
This makes no sense to me at all.

First show that it is true for n=1.
[a^1]^5 = a^5
a^(5*1) = a^5
So it is true for n=1.
Now let's show that if it is true for the integer k then it is also true for k+1.
[a^(k+1)]^5
= [a^k * a]^5
= [a^k]^5 * a^5
= a^(5k) * a^5 because we said it is true for k
= a^(5k+5)
= a^[5(k+1)]
so it is true for k+1.
So, by induction, if it is true for 1, and also if it is true for a certain integer then it is also true for the next integer, it must be true for all positive integers.
[a^1]^5 = a^5
a^(5*1) = a^5
So it is true for n=1.
Now let's show that if it is true for the integer k then it is also true for k+1.
[a^(k+1)]^5
= [a^k * a]^5
= [a^k]^5 * a^5
= a^(5k) * a^5 because we said it is true for k
= a^(5k+5)
= a^[5(k+1)]
so it is true for k+1.
So, by induction, if it is true for 1, and also if it is true for a certain integer then it is also true for the next integer, it must be true for all positive integers.

Induction involves two steps. First you have to prove the statement is true for the base. Since n is in the positive integers, we start with n=1.
(a^1)^5 = a^5 = a^((5)(1)) = a^(5n)
So it is true for the base.
Step 2 is to assume the statement is true for 'n', and from that show that it is true for 'n+1'.
(a^n)^5 = a^(5n) < Assume that's true.
Now stick in 'n+1' and see what happens.
[a^(n+1)]^5 =
[(a^n)(a^1)]^5 =
[(a^n)(a)]^5
[(a^n)^5](a^5) =
[a^(5n)](a^5) = < Substituting what we assumed was true for (a^n)^5
a^(5n+5) =
a^(5(n+1))
And thus we have proven by mathematical induction that (a^n)^5) = a^(5n) is true for all n.
(a^1)^5 = a^5 = a^((5)(1)) = a^(5n)
So it is true for the base.
Step 2 is to assume the statement is true for 'n', and from that show that it is true for 'n+1'.
(a^n)^5 = a^(5n) < Assume that's true.
Now stick in 'n+1' and see what happens.
[a^(n+1)]^5 =
[(a^n)(a^1)]^5 =
[(a^n)(a)]^5
[(a^n)^5](a^5) =
[a^(5n)](a^5) = < Substituting what we assumed was true for (a^n)^5
a^(5n+5) =
a^(5(n+1))
And thus we have proven by mathematical induction that (a^n)^5) = a^(5n) is true for all n.

i guess:
first n=1
[a^1]^5=a^(5*1) true a=a
then n=k
[a^k]^5=a^(5k)
when n = k+1
[a^ (k+1)]^5=a^(5(k+1))
(a^k * a^1)^5 = (a^k)^5 * (a^1)^5 from previous expression we know that [a^k]^5=a^(5k)
substitute a^(5k) instead of (a^k)^5
a^(5k) * (a^1)^5 = a^(5k) * (a^5)^1 = a^(5k) * (a^5) = a^(5k + 5) = a^(5(k+1))
first n=1
[a^1]^5=a^(5*1) true a=a
then n=k
[a^k]^5=a^(5k)
when n = k+1
[a^ (k+1)]^5=a^(5(k+1))
(a^k * a^1)^5 = (a^k)^5 * (a^1)^5 from previous expression we know that [a^k]^5=a^(5k)
substitute a^(5k) instead of (a^k)^5
a^(5k) * (a^1)^5 = a^(5k) * (a^5)^1 = a^(5k) * (a^5) = a^(5k + 5) = a^(5(k+1))

[a^n]^5 = (a^n)*(a^n)*(a^n)*(a^n)*(a^n) = a^(n+n+n+n+n) = a^5n
LOL idk what induction is.
LOL idk what induction is.