I was reading the Tau Manifesto (no offence to pi fans) and realized you could do as follows. Starting with the Euler identity for a full rotation: $$e^{iτ}=1$$ If $e+τ=\frac{p}{q}$ then: $$e^{i(p/q-e)}=1$$ $$e^{ip/q}=e^{ie}$$ $$i\frac{p}{q}=ie$$ $$\frac{p}{q}=e$$ Which we know is false, therefore $e+τ$ must be irrational. Is there any flaw in this proof? I need to know!
EDIT A possible objection is that if $e^{im}=e^{in}$ in general then since $e^{iτ}=e^{0i}$, $τ=0$. In the Euler equation we are talking about rotations, and a rotation of $τ$ is equivalent to a rotation of $0$.
EDIT1 The resolution to this is that $e^{ia/b} = e^{ie}$ decomposes to: $$\frac{a}{b}+m\tau=e+n\tau$$ $$\frac{a}{b}=e+\tau(n-m)$$ $$\frac{a}{b}=e+k\tau$$ since m and n are just integers (the information that $k=1$ having been lost). Have accepted mweiss' answer since he got close.
As others have already noted, the complex exponential function is not one-to-one; specifically, since $e^{\tau i}=1$, for any $a, b$ with $b = a + n\tau$ for some integer $n$, we would have $e^{ai} = e^{bi}$. Therefore, if $e^{ai}=e^{bi}$ then the most we can conclude is that $ai = bi + n \tau i$ for some $n$.
In your proof, then, the argument would run like this:
If $e+\tau=\frac{p}{q}$ then: $$e^{i(p/q-e)}=1$$ $$e^{ip/q}=e^{ie}$$ $$i\frac{p}{q}=ie + n\tau i$$ $$\frac{p}{q}=e + n\tau$$
So the conclusion is that if $e + \tau$ is rational, then $e + n\tau$ is rational for some $n$. But we knew that already.