Does This Proof Prove the Equation (a^{n})^{m}=a^{nm}?

[a_skier]

Homework Statement

Prove (a^{n})^{m}=a^{nm}

Homework Equations

Proof by induction
a^{n}*a=a^{n+1}
a^{n}*a^{m}=a^{nm}

The Attempt at a Solution

Let a and n be fixed. I will induct on m.

Suppose m=1. Then a^{(n)(m)}=a^{n(1)}=(a^{n})^{1}

Now assume the hypothesis is true for any integer m in P. I will show this is true for m+1.

a^{n(m+1)}=(a^{n})^{m+1}

Thus the hypothesis is true for m+1.

Is this proof sufficient? I am once again struck by a problem that seems almost too simple to be proved.

 

[Office_Shredder]

Your two equations in section 2 of your post seem fairly irrelevant to the problem at hand.

Your inductive step seems to have zero content. Why is
a^{n(m+1)} = \left( a^{n} \right)^{m+1}
true at all? The whole point is you're supposed to prove that using your inductive hypothesis. You need to use the fact that a^{nm} = \left(a^n\right)^m
at some point

 

[Fightfish]

Your induction will still only prove that the relationship is true for integer m and n if you succeed. Clearly the relationship is more general than that.

 

[a_skier]

a^n(m+1)=a^n*(m+1)=(by the theorem) (aⁿ)^(m+1)

How else could I move from the left side of the equality to the right? If someone could give me a little direction I would love to solve it myself.

 

[a_skier]

Your induction will still only prove that the relationship is true for integer m and n if you succeed. Clearly the relationship is more general than that.

Luckily I only need to prove it for integers right now. I don't think I could handle anything more atm.

 

[Fightfish]

You need to show explicitly that the proposition is true for m+1 making use of the assumption that the proposition holds true for m, which you are clearly lacking here.
Start with (a^{n})^{(m+1)}, rearrange / rewrite it until you manage to obtain a term (a^{n})^{m} somewhere. Then you can replace that term with a^{nm}

 

[Office_Shredder]

a^n(m+1)=a^n*(m+1)=(by the theorem) (aⁿ)^(m+1)

How else could I move from the left side of the equality to the right? If someone could give me a little direction I would love to solve it myself.

Fightfish gives a good suggestion... alternatively from a^n*(m+1) do literally the only algebra you are allowed to do and re-write it as a^nm+n

 

[a_skier]

Here's the proof for m=1

(aⁿ)^m=(aⁿ)¹=(By the definition)aⁿ

Now for m+1
Because x^n+m=xⁿ*x^m

Let x=(aⁿ)

(x)^(m+1)=(x)^m*(x)¹ and by the hypothesis and the case m=1 I have already proved and replacing x with (aⁿ)
(x)^m*(x)¹=(aⁿ)^m*(aⁿ)¹=a^nm*aⁿ=a^nm+n=a^n(m+1)

I'm pretty sure this proves it!? IM SO EXCITED THIS IS AWESOME HAHA!

 

[Office_Shredder]

Yeah that looks good to me