= Solution
The <Liouville approximation theorem> says that if $\alpha$ is an irrational <algebraic number> of degree $d$, there is $c_\alpha>0$ such that $|\alpha-p/q|\ge c_\alpha q^{-d}$ for all integers $p$ and $q\ge1$. To see the source of the bound, a degree-$d$ integer <minimal polynomial> $P$ gives $q^dP(p/q)$ a nonzero integer. On a fixed neighbourhood of $\alpha$, the <mean value theorem> bounds $|P(p/q)|$ by a constant times $|p/q-\alpha|$; outside that neighbourhood a smaller uniform constant suffices.
Let $\alpha$ denote the factorial-denominator sum and let $p_N/q_N$ be its partial sum through $N$, with $q_N=10^{N!}$ for $N\ge2$. This is an integer numerator even though $0!=1!=1$ makes the first two terms equal. Positivity and the rapid increase of factorial exponents imply
$$
0<\alpha-\frac{p_N}{q_N}<2\,10^{-(N+1)!}=2q_N^{-(N+1)}.
$$
For example the first omitted term is $10^{-(N+1)!}$ and every succeeding factorial exponent increases by at least one, giving a geometric tail bound less than $10/9$ times that first term.
First $\alpha$ is irrational: if $\alpha=a/b$, the positive difference from a partial sum is at least $1/(bq_N)$, contradicting this bound for sufficiently large $N$. If it were algebraic of any fixed degree $d$, <Liouville approximation theorem> would require $c_\alpha q_N^{-d}<2q_N^{-(N+1)}$, again impossible for large $N$. Hence $\boxed{\alpha\text{ is transcendental}.}$
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