The standard approach to this problem is to write $\xi(s) = \dfrac{s(s-1)}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s).$ This function is entire, and has zeroes precisely at the non-trivial zeroes of $\zeta(s)$. It also has a slow enough rate of growth that it can be written as a product over its zeroes: $$\xi(s) = \xi(0) \prod_{\rho}(1-\dfrac{s}{\rho}),$$ where the product is over zeroes of $\xi(s)$, i.e. over non-trivial zeroes of $\zeta(s)$. (Here I am following more-or-less the notation in Edward's book Riemann's zeta function, which is a good reference for these sort of things.) [Edit: Also, to ensure convergence, the product should be taken over "matching pairs" of zeroes, i.e. the factors for a pair of zeroes of the form $\rho$ and $1-\rho$ should be combined.] We can then write $$\zeta(s) = \dfrac{2\xi(0)}{s(s-1)}\pi^{s/2}\dfrac{1}{\Gamma(s/2)}\prod_{\rho}(1-\dfrac{s}{\rho}).$$ If we now replace $\dfrac{1}{\Gamma(s/2)}$ by its Weierstrass product, we get a product formula for $\zeta(s)$, namely $$\zeta(s) = \dfrac{\xi(0)}{s-1} (\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1 +\dfrac{s}{2n})e^{-s/2n} \prod_{\rho}(1-\dfrac{s}{\rho}).$$ (Here $\gamma$ is Euler's constant.)
Note that we can now compute $\xi(0)$, because we know that the value of $\zeta(s)$ at $s = 0$ is equal to $-1/2$. We find that $\xi(0) = 1/2$, and so $$\zeta(s) = \dfrac{1}{2(s-1)}(\pi e^{\gamma})^{s/2}\prod_{n=1}^{\infty}(1+\dfrac{s}{2n})e^{-s/2n}\prod_{\rho}(1-\dfrac{s}{\rho}).$$
An added cultural remark: Riemann's explicit formula for the prime counting function is obtained by taking a Fourier transform of the logarithm of this formula. Combined with the fact that all the non-trivial zeroes $\rho$ have real part $< 1$ (proved by Hadamard and de la Vallee Poussin), this gives the prime number theorem.