Need someone to show me how the Zeta function is equal to Euler's product formula?

Question

$\zeta(x)$=$1+\frac{1}{2^x}+\frac{1}{3^x}+\frac{1}{4^x}+\frac{1}{5^x}+\frac{1}{6^x}+\cdots$
Euler's product formula states, for all $x$ greater than $1$ we have:
$\zeta(x)$=$1\over 1-\frac{1}{2^x}$$\times$$1\over 1-\frac{1}{3^x}$$\times$$1\over 1-\frac{1}{5^x}$$\times$$1\over 1-\frac{1}{7^x}$$\times$$1\over 1-\frac{1}{11^x}$$\times \cdots$
The right hand side of the equation is an infinite product of the form $1\over 1-\frac{1}{p^x}$ where $p$ runs through all prime numbers. Also the author states that if we think of the infinite product as a limit, it will get closer to the limit value and that is the Zeta function itself.
Can someone show me a proof of this:
$\zeta(x)$=$1\over 1-\frac{1}{2^x}$$\times$$1\over 1-\frac{1}{3^x}$$\times$$1\over 1-\frac{1}{5^x}$$\times$$1\over 1-\frac{1}{7^x}$$\times$$1\over 1-\frac{1}{11^x}$$\times \cdots$