= Solution
Let
$$
\alpha_N=\sum_{n=1}^N2^{-n!}=\frac{p_N}{q_N},
\qquad q_N=2^{N!}.
$$
The tail satisfies
$$
0<\alpha-\alpha_N<2\,2^{-(N+1)!}.
$$
For every fixed positive <integer> $d$, this is smaller than $q_N^{-d}$ for all sufficiently large $N$.
The binary expansion has $1$ digits at factorial positions and arbitrarily long blocks of zeros, but it is neither eventually zero nor eventually periodic. Since a <rational number> has an <eventually periodic positional expansion of a rational number>[eventually periodic binary expansion], $\alpha$ is <irrational number>[irrational]. If it were an <algebraic number> of degree $d$, part (b), or equivalently the <Liouville approximation theorem>, would give a fixed positive multiple of $q^{-d}$ as a lower bound for all rational approximations. The fractions $p_N/q_N$ violate that bound. Thus $\alpha$ is a <transcendental number>.
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