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According to the fundamental theorem of arithmetic (unique factorization theorem), you can write every number as the product of some prime numbers, for example $33 = 11 \cdot 3$.

However, how can you do this when you're dealing with a prime number? If you write $29 = 29 \cdot 1$ you use 1 and that isn't a prime number. Should you just write $29 = 29^1$?

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    $\begingroup$ yes, every prime $p$ is just $p^1$. $\endgroup$ – Ittay Weiss May 21 '13 at 20:23
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    $\begingroup$ The fun question is: what is the prime factorization of $1$? $\endgroup$ – Qiaochu Yuan May 21 '13 at 20:24
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    $\begingroup$ @QiaochuYuan Any number to the zeroth? But that wouldn't be unique.. $\endgroup$ – Primesss May 21 '13 at 20:25
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    $\begingroup$ @QiaochuYuan, "product of 0 primes" is good enough for me. $\endgroup$ – vonbrand May 21 '13 at 20:26
  • $\begingroup$ From an algebraic point of view, $1$ doesn't admit a prime factorization :) When you look up the definitions of primes and unique factorization domains, you see that only non-invertible ring elements have a prime factorization. $1$ has the rare property of being an invertible integer, so it doesn't get to have a prime factorization. $\endgroup$ – Yoni Rozenshein May 22 '13 at 22:25
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A single number, like $31$ or $7$, is in fact a product as far as mathematics is concerned. It is a product of $1$ integer.

Indeed, you can even have a product of $0$ integers. This is defined to be $1$, because $1$ is the identity element of multiplication. (See Qiaochu's comment.)

When we say that an integer has a unique prime factorization, we mean it can be written as a product of some nonnegative integer number of primes. Thus, "$2 \cdot 2 \cdot 23$", "$31$", "$7$", and "$\quad$" are all valid prime factorizations.

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