First lets assume \(m \ge n\). For example, if \(m=5\) and \(n=2\), then:
\[ \frac{a*a*a*a*a}{a*a} = a*a*a\]
The 2 \(a\)'s in the denominator cancelled out the 2 \(a\)'s in the numerator. We can generalize this:
\[ \frac{\overbrace{a*a*\ldots}^{m}}{\overbrace{a*a*\ldots}^{n}} = \overbrace{a*a*\ldots}^{m-n}= a^{m-n} \]
What if \(m\) is less than \(n\)? For example if \(a^2/a^5\):
\[ \frac{a*a}{a*a*a*a*a} = \frac{1}{a*a*a}\]
All the \(a\)'s in the numerator cancelled outs ome of \(a\)'s in the denominator. We can generalize this:
\[ \frac{\overbrace{a*a*\ldots}^{m}}{\overbrace{a*a*\ldots}^{n}} = \frac{1}{\overbrace{a*a*\ldots}^{n-m}}\]
Since \(1/a^k = a^{-k}\). Then:
\[ \frac{\overbrace{a*a*\ldots}^{m}}{\overbrace{a*a*\ldots}^{n}} = \frac{1}{\overbrace{a*a*\ldots}^{n-m}} = a^{m-n}\]
we have proven that regardless of whether \(m\) or \(n\) is bigger, \(a^m ÷ a^n = a^{(m-n)}\). Keep in mind this proof only works when \(m\) and \(n\) are integers.