Scope of the method and open questions

The same mechanism elsewhere

Fourier's argument is a model of its kind: produce a quantity that must be an integer while lying strictly between $0$ and $1$. It reappears in several famous proofs.

Result	Author, date	Main tool
$e$ irrational	Euler, 1737	continued fraction
$e$ irrational	Fourier, around 1815	series and factorial
$\pi$ irrational	Lambert, 1761	continued fraction of $\tan$
$e$ transcendental	Hermite, 1873	Padé approximants
$\pi$ transcendental	Lindemann, 1882	extension of Hermite's work
$\pi$ irrational, short proof	Niven, 1947	integral of a well-chosen polynomial

The case of $e^2$ and powers of $e$

The previous argument does not carry over as is to $e^2$: the series $\sum \dfrac{2^k}{k!}$ cannot be cleared so simply, because the factor $2^k$ inflates the remainder. Liouville got around the obstacle in 1840: if we assume $e^2 = \dfrac{p}{q}$, we write

$$q\,e = p\,e^{-1}$$

then expand both sides as series, exploiting the alternating signs of $e^{-1} = \sum \dfrac{(-1)^k}{k!}$. The same contradiction appears.

More generally, a theorem of Lambert states that $e^r$ is irrational for every rational $r \neq 0$, and the Hermite–Lindemann theorem gives even more: $e^\alpha$ is transcendental for every algebraic number $\alpha \neq 0$.

Euler's continued fraction

Euler's historical approach relies on the expansion

$$e = [\,2 \,;\, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, \ldots\,]$$

where the pattern $1, 2m, 1$ repeats indefinitely. The criterion is clear-cut: a real number is rational if and only if its continued fraction expansion is finite. Since this expansion is infinite and perfectly regular, $e$ is irrational. As a bonus, this route gives excellent rational approximations of $e$, such as $\dfrac{2721}{1001}$.

What we still do not know

The frontier of knowledge is surprisingly close. It is still unknown whether $e + \pi$ is irrational, and likewise for $e \cdot \pi$ or $e^e$. We only know that they cannot both be algebraic: if $e + \pi$ and $e\,\pi$ were algebraic, then $e$ and $\pi$, roots of the polynomial

$$X^2 - (e + \pi)\,X + e\,\pi$$

would be algebraic too, which Hermite and Lindemann rule out. At least one of the two is therefore transcendental — but nobody knows which.

Classic pitfall

Believing that a sum or product of two irrational numbers is irrational. That is false: $\sqrt{2} + \left(1 - \sqrt{2}\right) = 1$ and $\sqrt{2} \times \sqrt{2} = 2$. This is exactly why the status of $e + \pi$ remains open, even though $e$ and $\pi$ are both transcendental.