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?

I figured out an elementary proof that $e^2$ is irrational, although it doesn't seem to generalize to other powers:
Suppose $e^2$ is rational.  Then there exist $a, b, k, n$ such that $e^2 \cdot 2^k = \frac{a}{b}$, $b = \frac{(2^n)!}{2^{2^n-1}}$, and $b$ is an odd integer.  By definition (Taylor series for $e^x$) we have $2^k \cdot e^2 = 2^k \cdot \sum_{j \ge 0}{\frac{2^j}{j!}} = H_n + T_n$, $H_n = 2^{k} \cdot \sum_{0 \le j \le 2^n}{\frac{2^{j}}{j!}}$, $T_n = 2^k \cdot \sum_{j \gt 2^n}{\frac{2^j}{j!}}$.  Since every term of the sum in the definition of $H_n$ can be reduced to an odd denominator, and $b$ is an integer divisible by every such odd denominator, $b \cdot H_n$ is an integer.  And $0 \lt b \cdot T_n \lt \frac{2^{k+1}}{2^n}$, so for sufficiently large $n$, $b \cdot T_n$ is not an integer.  But $2^k \cdot b \cdot e^2 = b \cdot H_n + b \cdot T_n = a$ is an integer, contradicting the assumption that $e^2$ is rational.
