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.

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

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