The Riemann $\zeta$ function plays a significant role in number theory and is defined by $$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \qquad \text{ for } \sigma > 1 \text{ and } s= \sigma + it$$
The Riemann hypothesis asserts that all the non-trivial zeroes of the $\zeta$ function lie on the line $\text{Re}(s) = \frac{1}{2}$.
My question is:
Why are we interested in the zeroes of the $\zeta$ function? Does it give any information about something?
What is the use of writing $$\zeta(s) = \prod_{p} \biggl(1-\frac{1}{p^s}\biggr)^{-1}$$
Best Answer
Short answer: Understanding the distribution of the prime numbers is directly related to understanding the zeros of the Riemann Zeta Function.**
Long Answer: The prime counting function is defined as $\pi(x)=\sum_{p\leq x} 1,$ which counts the number of primes less than $x$. Usually we consider its weighted modification $$\psi(x)=\sum_{p^{m}\leq x}\log p$$ where we are also counting the prime powers. It is not hard to show that $$\pi(x)=\frac{\psi(x)}{\log x}\left(1+O\left(\frac{1}{\log x}\right)\right),$$ which means that these two functions differ by about a factor of $\log x$.
The prime number theorem states that $\psi(x)\sim x$, but this is quite hard to show. It was first conjectured by Legendre in 1797, but took almost 100 years to prove, finally being resolved in 1896 by Hadamard and de la Vallée Poussin. In 1859 Riemann outlined a proof, and gave a remarkable identity which changed how people thought about counting primes. He showed that (more or less) $$\psi(x)=x-\sum_{\rho:\zeta(\rho)=0}\frac{x^{\rho}}{\rho}-\frac{\zeta^{'}(0)}{\zeta(0)},$$ where the sum is taken over all the zeros of the zeta function. ${}^{++}$
Notice that this is an equality. The left hand side is a step function, and on the right hand side, somehow, the zeros of the zeta function conspire at exactly the prime numbers to make that sum jump. (It is an infinite series whose convergence is not uniform) If you remember only 1 thing from this answer, make it the above explicit formula.
An equivalence to RH: Current methods allow us to prove that $$\psi(x)=x+O\left(xe^{-c\sqrt{\log x}}\right).$$ This error term decreases faster then $\frac{x}{(\log x)^A}$ for any $A$, but increases faster then $x^{1-\delta}$ for any small $\delta>0$. In particular, proving that the error term was of the form $O\left(x^{1-\delta}\right)$ for some $\delta>0$ would be an enormous breakthrough. The Riemann Hypothesis is equivalent to showing the error term is like square root $x$, that is proving the statement $$\psi(x)=x+O\left(x^{\frac{1}{2}}\log^{2}x\right).$$ In other words, the Riemann Hypothesis is equivalent to improving the error term when counting the prime numbers.
Remark: In your question you incorrectly state the Riemann Hypothesis, which says that all zeros have real part $\frac{1}{2}$. The fact that infinitely many zeros lie on the line was shown by Hardy in 1917, and in 1942 Selberg showed that a positive proportion lie on the line. In 1974 Levinson showed that this proportion was at least $\frac{1}{3}$, and Conrey 1989 improved this to $\frac{2}{5}$.
** Of course, there may be some people who are interested in the zeros of the zeta function for other reasons. Historically the prime numbers are what first motivated the study of the zeros.
${}^{++}$: Usually the trivial zeros will be separated out of the sum, but I do not make this distinction here. Also, Riemann's original paper states things in terms of $\Pi(x)$ and $\text{li}(x)$, the Riemann pi function and logarithmic integral, rather then $\psi(x)$. This is a very slight difference, and I use $\psi(x)$ above because it is easier and cleaner to do so.
See also: Why is $\zeta(1+it) \neq 0$ equivalent to the prime number theorem?