= Solution
<Liouville approximation theorem> states that if an irrational algebraic number $\alpha$ has degree $d\geq2$, then some $C(\alpha)>0$ satisfies
$$
\left|\alpha-\frac pq\right|>\frac{C(\alpha)}{q^d}
$$
for every rational $p/q$ with $q>0$.
The omitted $n=0$ term is rational and does not affect transcendence. Let
$$
\xi=\sum_{n\geq1}10^{-n^n},\qquad
\xi_N=\sum_{n=1}^N10^{-n^n}=\frac{p_N}{q_N},\quad q_N=10^{N^N}.
$$
The 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,
$$
0<\xi-\xi_N<2\,10^{-(N+1)^{N+1}}.
$$
For every fixed $d$, the exponent $(N+1)^{N+1}$ eventually exceeds $dN^N$ by an arbitrarily large amount, so this upper bound is smaller than $Cq_N^{-d}$. 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.
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