The idea of multiplication is just to simplify the marking of a sum. For example, [br][br]  [math]\LARGE 2+2+2+2+2+2+2+2+2+2[/math][br]  [br]is much easier to read in the form of [math]\Large 10\cdot 2[/math].  The powers have exactly the same idea: [math]\Large 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^6[/math] . [br][color=#0000ff][br]Example 1.[/color] Let us study the case [math]\Large 2\cdot 3\cdot 2\cdot 3\cdot 2\cdot 3\cdot 2\cdot 3[/math]. We can rewrite the expression in the form of [br][br]  [math]\Large 2\cdot 2\cdot 2 \cdot 2 \cdot 3\cdot 3\cdot 3 \cdot 3=2^4\cdot 3^4[/math][br][br]or in the form[br][br]  [math]\Large (2\cdot 3)\cdot (2\cdot 3)\cdot (2\cdot 3)\cdot (2\cdot 3)=(2\cdot 3)^4.[/math][br][br]As both are forms of the same original expression, we can say that[br][br]  [math]\Large 2\cdot 2\cdot 2 \cdot 2 \cdot 3\cdot 3\cdot 3 \cdot 3=2^4\cdot 3^4 =(2\cdot 3)^4.[/math][br][br][br][color=#0000ff]Example 2.[/color] How about [math]\Large \left ( 2^3 \right )^2[/math]?[br][br][math]\Large \left ( 2^3 \right )^2=2^3\cdot 2^3 =2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^6[/math].[br][br]The power in the last term is exactly same as the product of the powers in the original expression.[br][br][br]The formulas of powers are expressed as[br][br][math]\LARGE\textcolor{blue}{\begin{eqnarray}[br]a^ma^n&=&a^{m+n}\\[br]\frac{a^m}{a^n}&=&a^{m-n}\\[br](ab)^n&=&a^nb^n\\[br]\left (\frac{a}{b}\right )^n&=&\frac{a^n}{b^n}\\[br]\left ( a^m\right )^n&=&a^{mn}[br]\end{eqnarray}}[br][/math][br] [br]  [br]Based on these formulas, [math]\Large \textcolor{blue}{a^0=1}[/math].[br][br][color=#0000ff]Example 3.[/color] [br][br]  [math]\LARGE \frac{2^3\cdot 3^3}{36}=\frac{2^3\cdot 3^3}{4\cdot 9}=\frac{2^3\cdot 3^3}{2^2\cdot 3^2}=2^{3-2}\cdot 3^{3-2}=2\cdot 3=6[/math][br]   [br]The same with parameters:[br][br]  [math]\LARGE \frac{a^3\cdot b^3}{a^2\cdot b^2}=a^{3-2}\cdot b^{3-2}=a\cdot b[/math][br]              [br]   [br]              [br][br][color=#0000ff]Example 4[/color].[br][br]  [math]\LARGE \left ( \frac{8\cdot 27}{72}\right )^2 =\left (\frac{8\cdot 3^3}{8\cdot \underbrace{9}_{3^2}} \right )^2=(3^{3-2})^2=3^2=9[/math][br]  [br] [br][u]Problem 1[/u]. Can you explain with words, why [math]\Large a^{-n}=\frac{1}{a^n}[/math]?