Showing $\pi$ is irrational using taylor's theorem

Question

To prove the irrationality of $$e = \sum ^\infty _ {n=0} \frac{1}{n!}$$ we can show that $e \lt 3$ by using a suitable geometric series. By Taylor's theorem (applied to $a=0$ and $b=1$) we know that, $$ e-\sum ^n _{k=0} \frac{1}{n!} = R_n = e^\alpha \frac {1}{\bigl(n+1\bigl)!} $$ for some $\alpha$ such that $0 \lt \alpha \lt 1$. Since $e \lt 3$, $R_n \lt \frac{3}{(n+1)!}.$ Now suppose that $e$ is a rational number $\frac{c}d$ where $c$ and $d$ are co-prime. For $n=d$ we see that $d! R_d \in Z$. On the other hand, using the estimate for $R_d$ that we have obtained using Taylor's theorem, $$d! R_d \lt \frac{d! \cdot 3}{(d+1)}! \lt 1$$ if $d \geq 2$
How can I show $\pi$ is irrational using similar ideas?