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Euler's product is an infinite product over the prime numbers that has numerous applications in analysis and number theory. It is an alternative form of the Riemann zeta function

Where \mathrm{Re}(z) > 1, the Euler product is equal to,

\zeta(z) = \sum_{n \geq 1} \frac{1}{n^z} = \prod_{p | \text{prime}} \frac{1}{1-p^{-z}}

Where z is some complex number.

Sketch of a proof

Consider the following series,

\zeta(z) = 1 + \frac{1}{2^z} + \frac{1}{3^z} + \frac{1}{4^z} \ldots

Now, multiply through \frac{1}{2^{z}},

\frac{1}{2^z}\zeta(z) = \frac{1}{2^z} + \frac{1}{4^z} + \frac{1}{6^z} + \frac{1}{8^z} \ldots

By subtracting these two series, we will remove every other term, hence removing any with a denominator that divides 2.

\left(1 - \frac{1}{2^z}\right)\zeta(z) = 1 + \frac{1}{3^s} + \frac{1}{5^s} + \frac{1}{7^s} + \frac{1}{9^s} \ldots

One can repeat this process infinitely with every prime, p to remove every pth term, in turn removing any number with a denominator divisible by p. This mimics the behaviour of the Sieve of Eratosthenes. Notice that this will remove all primes, leaving only 1, as one is not conventionally considered prime and is excluded in the Sieve of Eratosthenes for obvious technical reasons. This forms an infinite series on the LHS while leaving 1 on the RHS,

\left(\prod_{p | \text{prime}} 1 - p^{-z}\right)\zeta(z) = 1.

Dividing both sides, we can obtain,

\zeta(z) = \prod_{p | \text{prime}} \frac{1}{1-p^{-z}}
\blacksquare

Consequences regarding the infinitude of prime numbers

Let z=1, we know that

\zeta(1) = \prod_{p | \text{prime}} \frac{1}{1-p^{-1}}

(ignore technicalities regarding that equality statement for the time being)

Observe that \zeta(1) represents the harmonic series, which quite famously diverges. As it diverges, hence does the RHS. As the product on the RHS also diverges, that implies that there must be an infinite number of terms, else the product would converge. This implies that there is an infinite number of primes. \blacksquare

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