# Proof of (a^n)^m=a^(nm)

1. Jun 17, 2013

### a_skier

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Jun 17, 2013

### Office_Shredder

Staff Emeritus
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

3. Jun 17, 2013

### 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.

4. Jun 17, 2013

### a_skier

an(m+1)=an*(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.

5. Jun 17, 2013

### a_skier

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

6. Jun 17, 2013

### 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 $a^{nm}$

7. Jun 17, 2013

### Office_Shredder

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

8. Jun 17, 2013

### a_skier

Here's the proof for m=1

(an)m=(an)1=(By the definition)an

Now for m+1
Because xn+m=xn*xm

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=anm*an=anm+n=an(m+1)

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

9. Jun 17, 2013

### Office_Shredder

Staff Emeritus
Yeah that looks good to me