Proving that if $a^n=b^n$, then $a=b$, for positive $a$ and $b$, and natural $n$ In high school algebra, natural number exponents are defined as
$$\begin{aligned}a^1 &= a\\
a^{(n + 1)} &= a^n a\end{aligned}$$
With these I used induction to prove the 3 exponent laws. I also proved some ancillary theorems such as the nth power of a positive is positive, distribution of powers over fractions, and so on. Then I got to this statement:

For positive $a$, $b$ and natural number $n$,
$$\text{if }a^n = b^n\text{ then } a = b\text{.}$$

I have a proof, but it is very different from any of the other proofs. It requires $\Sigma$ notation, which none of the others require. However, I can't figure out how to do it without $\Sigma$. Is it possible?
Here's my proof:
$$a^n = b^n\\
  a^n - b^n = 0\\
  (a - b) \left(\sum_{k=1}^n \frac{a^n b^k}{a^k b}\right) = 0\\
  a = b  \quad\text{or}\quad \sum_{k=1}^n \frac{a^n b^k}{a^k b} = 0$$
As the summation has at least $1$ term and every term is positive, the summation must be positive, which is not equal to $0$. This leaves $a = b$.
Q.E.D.
 A: Prove the contrapositive: if $a\ne b$, $a^n\ne b^n$.
If $a\ne b$ then without loss of generality $a>b$. Now prove by induction $a^n>b^n$. The inductive step boils down to $a^{k+1}>a^kb>b^{k+1}$.
A: Your idea will work if you get the summations right. As an alternative, you could prove by induction that, for positive $x$ and $y$, if $x > y$, then $x^n > y^n$. Given that, if $a^n = b^n$, then $a < b$ and $b < a$ are both impossible, so you must have $a = b$.
A: Here's another approach.
Lemma If $a$ and $b$ are real numbers such that $0 < a < b$, and $n$ is a positive integer, then $0 < a^n < b^n$.
You should try to prove this by induction!
Claim If $a$ and $b$ are positive real numbers and $n$ is a positive integer such that $a^n = b^n$, then $a = b$.
Proof. Let $a$ and $b$ be positive real numbers, and let $n$ be a positive integer. We proceed by contrapositive: suppose $a \neq b$. Then $0 < a < b$ or $a > b > 0$. By the lemma, this yields $0 < a^n < b^n$ or $a^n > b^n > 0$. In both cases, $a^n \neq b^n$.
A: You could write it explicitly. Suppose $a,b$ are not-negative real numbers and $n$ is a positive natural number. Is easy to show that
\begin{align}
a^n - b^n  
= (a-b)( a^{n-1} + a^{n-2}b^{1} + a^{n-3}b^{2} + ... + a^{1}b^{n-2} + b^{n-1})
\end{align}
This expression should be familiar from geometric sums and can be seen by distributing the first factor,  most of the term cancel out. From this, the second factor is not-negative,
$$  a^{n-1} + a^{n-2}b^{1} + a^{n-3}b^{2} + ... + a^{1}b^{n-2} + b^{n-1} \geq 0 \; , $$
since it is a sum of positive terms. So, the same way you proved,
\begin{align}
0 &= a^n - b^n \, \\[5pt] \Rightarrow \;  0&=a-b \\[5pt]
 \Rightarrow \; a & = b \; . \qquad \qquad \blacksquare
\end{align}

Note that you need $n$ to be is nonzero and that $a,b$ to be non-negative. For example $(2)^2=(-2)^2$ but $2 \neq -2$, well as $x^0 = 1 $ for any $x$.
A: Direct Proof
Here, we will prove a stronger result:

*

*If, $a^n=b^n$ and $a,b>0, n \in \mathbb {R} \setminus \left\{0\right\}$, then we have

$$\ln a^n =\ln b^n$$
$$\implies n \ln a=n \ln b$$
$$\implies \ln a=\ln b$$
$$\implies \ln a-\ln b=0$$
$$\implies \ln \left(\frac ab\right )=0$$
$$ \implies \dfrac ab =e^0=1$$
$$\color {gold}{\boxed {\color{black}{\dfrac ab=1 \Longrightarrow a=b.}}}$$
