Is there a simple proof that $e^2$ is irrational using a positional numeral system?

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.

@Robert Israel was very close.

Assume $e^2=p/q$. Rewrite as $qe-pe^{-1}=0$, or $$2q+{q-p\over2!}+{q+p\over3!}+{q-p\over4!}+\cdots=0$$ This isn't exactly in the form of the expansion in the first paragraph of the question, but the numerators are bounded in absolute value, which is more than enough to make the proof go through.

Not quite $e^2$, but consider $x = a e + b/e$ for integers $a,b$ with $|b| \le a$. Then $x = \sum_{j=0}^\infty c_j/j!$ where $c_j = a+b$ when $j$ is even and $a-b$ when $j$ is odd. Write $x = r + y$ where $r = \sum_{j = 0}^{a+b} c_j/j!$ is rational and $y = \sum_{j=a+b+1}^\infty c_j/j!$ is a non-terminating expansion in factorial base, and thus is irrational. Therefore $x$ is irrational.