Areteem: Response to a question from one of the students in Number Theory II-A Self-paced course

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Response to a question from one of the students in Number Theory II-A Self-paced course

One student asked the following question to the teacher:

Just had a question about the proof of Fermat's Little Theorem. In the first step, you said that in mod $p$ , $(p-1)!\cdot a^{p-1}$ equals $1a\cdot 2a\cdot 3a\cdots (p-1)a$. But, why wouldn't the $p-1$ exponent on $a$ carry through? Wouldn't it be equal to $1 a^{p-1}\cdot 2a^{p-1}\cdots(p-1)a^{p-1}$?

To understand this, we need to understand exactly what $(p-1)!\cdot a^{p-1}$ means: it is just a bunch of numbers multiplied together. $(p-1)!$ equals $1\cdot 2\cdot 3\cdots (p-1)$, and $a^{p-1}$ equals $a\cdot a\cdot a\cdots a$ (there are $p-1$ copies of $a$). Multiplication has the commutative property, so if we rearrange the order of the multiplication, we can get $1\cdot a\cdot 2\cdot a\cdot 3\cdot a\cdots (p-1)\cdot a$. In this case, the exponent $p-1$ indicates how many $a$'s are being multiplied, so once we split all the $a$'s individually, no exponents are needed for each of the $a$'s.

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