What is the simplest way to prove that the logarithm of any prime is irrational? What is the simplest way to prove that the logarithm of any prime is irrational?
I can get very close with a simple argument: if $p \ne q$ and $\frac{\log{p}}{\log{q}} = \frac{a}{b}$, then because $q^\frac{\log{p}}{\log{q}} = p$, $q^a = p^b$, but this is impossible by the fundamental theorem of arithmetic.  So the ratio of the logarithms of any two primes is irrational.  Now, if $\log{p}$ is rational, then since $\frac{\log{p}}{\log{q}}$ is irrational, $\log{q}$ is also irrational.  So, I can conclude that at most one prime has a rational logarithm.
I realize that the rest follows from the transcendence of $e$, but that proof is relatively complex, and all that's left to show is that no integer power of $e$ is a prime power (because if $\log p$ is rational, then $e^a = p^b$ has a solution).  It is easy to prove that $e$ is irrational ($e = \frac{a}{b!} = \sum{\frac{1}{n!}}$, multiply by $b!$ and separate the sum into integer and fractional parts) but I can't figure out how to generalize this simple proof to show that $e^x$ is irrational for all integer $x$; it introduces a $x^n$ term to the sum and the integer and fractional parts can no longer be separated.  How to complete this argument, or what is a different elementary way to show that $\log{p}$ is always irrational?

A proof of the irrationality of rational powers of $e$ is given on page 8 of Keith Conrad's notes. 
