 Solution

Liouville approximation theorem states that if an irrational algebraic number $\alpha$ has degree $d\ge 2$, then some $C(\alpha )>0$ satisfiesfor every rational $p/q$ with $q>0$.

The omitted $n=0$ term is rational and does not affect transcendence. LetThe decimal expansion has ones at the increasingly separated positions $n^n$ and zeros elsewhere, so it is not eventually periodic and $\xi$ is irrational. Moreover,For every fixed $d$, the exponent $(N+1{)}^{N+1}$ eventually exceeds $d{N}^N$ by an arbitrarily large amount, so this upper bound is smaller than $C{q}_N^{-d}$. Liouville's theorem therefore rules out every finite algebraic degree, proving that $\xi$ is transcendental.

There are only countably many integer polynomials and each has finitely many roots, so the algebraic numbers are countable. Since $\mathbb{R}$ is uncountable, its complement, the set of transcendental numbers, is uncountable.