Can the golden ratio accurately be expressed in terms of $e$ and $\pi$

The following is exact. :-)

$$\phi=\frac{\frac{\pi}{\pi}+\sqrt{\frac{e+e+e+e+e}{e}}}{\frac{e}{e}+\frac{\pi}{\pi}}$$

$e$ and $\pi$ are transcendental numbers, that is to say they are not the solution of any polynomial with rational coefficients. It's not hard to see that if $x$ is transcendental, then the following are also transcendental:

• $x \pm c$ for any rational number $c$,
• $kx$ for any nonzero rational number $k$ (so $x/k$ too),
• $x^n$ for any whole number $n > 1$,
• $\sqrt{x}$ and indeed $\sqrt[n]{x}$ for any whole number $n > 1$.

$\phi$ is not transcendental: it is the solution to a simple quadratic polynomial, so it won't be the result of any operations like the above applied to a transcendental number. The only way you're going to get an expression exactly equal to $\phi$ using only one of $e$ and $\pi$ and the above operations is by not really using them at all, e.g. cancelling them out completely, as other answers do.

What if you use both $e$ and $\pi$? Well, somewhat absurdly, it's not even known if $e + \pi$ is irrational, let alone transcendental! So with our current level of knowledge we can't say whether you can make $\phi$ exactly in a nontrivial way. However, I think most mathematicians would be extremely surprised if it turned out that $e + \pi$ or anything like it (apart from $e^{i\pi}$ and its family, of course!) were not transcendental, and hence mostly useless for constructing something like $\phi$.

At the time of writing, three of the other answers simply express the golden ratio by using expressions like $e/e$ and $\pi/\pi$ to get small integers. The fourth and final one discusses why a good solution is unlikely.

I believe using the imaginary unit $i=\sqrt{-1}$ results in the following very elegant solution: $$ \varphi = e^{i\pi/5} + e^{-i\pi/5}. $$

Edit: robjohn notes that one can directly derive the fundamental identity for the golden ratio $\varphi^2 = \varphi + 1$ from this expression:

$$ \begin{align} \color{#C00000}{\left(e^{i\pi/5}+e^{-i\pi/5}\right)}^2 &=e^{i2\pi/5}+2+e^{-i2\pi/5}\\ &=\left(e^{i2\pi/5}+1+e^{-i2\pi/5}\right)+1\\ &=-\left(e^{i4\pi/5}+e^{-i4\pi/5}\right)+1\\ &=\color{#C00000}{\left(e^{i\pi/5}+e^{-i\pi/5}\right)}+1 \end{align}. $$