You know from whole-number exponents that an exponent counts factors: $2^{4} = 2 \cdot 2 \cdot 2 \cdot 2$. That one idea is all you need to multiply powers. There is nothing new to memorize, only something to count.
Why it works
$x^{3} \cdot x^{4}$ means three factors of $x$ followed by four more: $(x \cdot x \cdot x) \cdot (x \cdot x \cdot x \cdot x)$. Multiplication can be regrouped any way you like, so this is just seven factors of $x$ in a row, which is $x^{7}$. The exponents are only counting, so they add.
Key idea
When you multiply powers with the same base, keep the base and add the exponents: $$a^{m} \cdot a^{n} = a^{m+n}$$
Worked examples
Worked example: Count the factors
Write $5^{2} \cdot 5^{6}$ as one power.
- $5^{2}$ is 2 factors of 5, and $5^{6}$ is 6 more.
- Together that is $2 + 6 = 8$ factors of 5.
- The base stays 5: $5^{2} \cdot 5^{6} = 5^{8}$.
Answer: $5^{8}$
Worked example: A hidden exponent
Simplify $y^{5} \cdot y \cdot y^{3}$.
- A plain $y$ is one factor, so $y = y^{1}$.
- Add all the exponents: $5 + 1 + 3 = 9$.
Answer: $y^{9}$
Worked example: Numbers in front
Simplify $(4a^{3})(-2a^{5})$.
- Multiplication can happen in any order, so group the numbers and the powers: $\big(4 \cdot (-2)\big)\big(a^{3} \cdot a^{5}\big)$.
- The numbers multiply: $4 \cdot (-2) = -8$.
- The powers of $a$ add exponents: $a^{3+5} = a^{8}$.
Answer: $-8a^{8}$
Common mistake
Writing $2^{3} \cdot 2^{4} = 4^{7}$ (multiplying the bases) or $2^{12}$ (multiplying the exponents). Neither one matches the count of factors. Check with real numbers: $2^{3} \cdot 2^{4} = 8 \cdot 16 = 128$, and $2^{7} = 128$. But $4^{7} = 16{,}384$ and $2^{12} = 4{,}096$.
The rule needs the same base. $2^{3} \cdot 3^{4}$ can’t become one power, because the factors are different numbers. In $x^{2}y^{3}$ the letters are different, so nothing combines.