Prove that if $x$ is a non-zero rational number, then $\tan(x)$ is not a rational number and use this to prove that $\pi$ is not a rational number.

Read the excellent book of Ivan Niven, "Irrational Numbers". There is the answer for your interesting question.

The proof from a few hundred years ago was done by Lambert and Miklós Laczkovich provided a simplified version later on. The Wikipedia page for "Proof that $\pi$ is irrational" provides this proof (in addition to some other discussion).

http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational#Laczkovich.27s_proof

Edit: Proving the more general statement here hinges upon Claim 3 in Laczkovich's proof. Defining the functions $f_k(x)$ by \begin{equation} f_k(x) = 1 - \frac{x^2}{k} + \frac{x^4}{2!k(k+1)} - \frac{x^6}{3!k(k+1)(k+2)} + \cdots \end{equation} it can be seen (using Taylor series) that \begin{equation} f_{1/2}(x/2) = \cos(x) \end{equation} and \begin{equation} f_{3/2}(x/2) = \frac{\sin(x)}{x} \end{equation} so that \begin{equation} \tan x = x\frac{f_{3/2}(x/2)}{f_{1/2}(x/2)} \end{equation}

Taking any $x \in \mathbb{Q} \backslash \{0\}$ we know that $x/2 \in \mathbb{Q} \backslash \{0\}$ and also that $x^2/4 \in \mathbb{Q} \backslash \{0\}$ as well. Then $x/2$ satisfies the hypotheses required by Claim 3.

Using Claim 3 and taking $k = 1/2$, we have \begin{equation} \frac{f_{k+1}(x/2)}{f_k(x/2)} = \frac{f_{3/2}(x/2)}{f_{1/2}(x/2)} \notin \mathbb{Q} \end{equation} which then also implies that \begin{equation} \frac{x}{2}\frac{f_{3/2}(x/2)}{f_{1/2}(x/2)} \notin \mathbb{Q} \end{equation} Multiplying by 2 then gives $\tan x \notin \mathbb{Q}$.