TITLE: Is there a simple proof that $e^2$ is irrational using a positional numeral system?
QUESTION [21 upvotes]: My favorite proof that $e$ is irrational goes something like this.  Observe that we can write any real number $r$ as
$$
a \,+\, \frac{b_2}{2} \,+\, \frac{b_3}{3!} \,+\, \frac{b_4}{4!} \,+\, \frac{b_5}{5!} \,+\, \cdots
$$
where $a\in\mathbb{Z}$ and each $b_n\in\{0,1,\ldots,n-1\}$.  This is the expansion of $r$ in the factorial number system, where $b_n$ is the $n$'th "digit". In particular,
$$
a \;=\; \lfloor r\rfloor\qquad\text{and}\qquad b_n = \big\lfloor n!\,(r-s_{n-1})\big\rfloor
$$
for each $n$, where $s_n$ denotes the $n$'th partial sum of the above series.
It is easy to see that $r$ is rational if and only if this expansion terminates.  Then $e$ must be irrational, since its expansion does not terminate:
$$
e \;=\; 2 \,+\, \frac{1}{2} \,+\, \frac{1}{3!} \,+\, \frac{1}{4!} \,+\, \frac{1}{5!} \,+\, \cdots
$$
Question: Can this proof somehow be modified to show that $e^2$ is irrational? It doesn't work straight off, since $2^n$ isn't in the range $\{0,1,\ldots,n-1\}$.
I'm also curious whether there are any other cases in which irrationality of an interesting number can be proven using a non-standard positional numeral system.

REPLY [3 votes]: The usual elementary proof that $e$ is irrational works also for $e^2$. 
Suppose that $e^2$ is rational. 
We know that 
$$ e^2 = \sum_{k=0}^\infty \frac{2^k}{k!}. $$
Let $q\ge2$ be a positive integer, and let $Q=2^q$. 
If $q$ is sufficiently large then $\dfrac{Q!}{2^{Q-q+1}}\cdot e^2$ is an integer. (Notice that the exponent of $2$ in the prime factorization of $\dfrac{Q!}{2^{Q-q+1}}$ is $q-2$.)
Now consider 
$$
\frac{Q!}{2^{Q-q+1}}\cdot e^2
-\sum_{k=0}^Q \frac{Q!}{2^{Q-q}} \cdot \frac{2^{k-1}}{k!}
=\sum_{k=Q+1}^\infty \frac{2^{k-Q-1}Q}{(Q+1)(Q+2)\ldots k} .
$$
The left-hand side is an integer and the right-hand side is in $(0,1)$.
Contradiction.