= Solution
The <Liouville approximation theorem> states that an irrational <algebraic number> $\alpha$ of degree $d\ge2$ satisfies $|\alpha-a/b|\ge C_\alpha b^{-d}$ for all integers $a$ and $b\ge1$, for some $C_\alpha>0$. The constant is independent of the rational approximation.
The required condition is \b[infinitely many of the coefficients equal one]. If only finitely many do, the <series> is a finite sum of <rational numbers>. Conversely let $\alpha$ be the sum with infinitely many nonzero terms. Its truncation $s_N$ has denominator dividing $q_N=10^{N!}$. For $N\ge2$, its strictly positive tail satisfies
$$
0<\alpha-s_N\le\sum_{j=N+1}^\infty10^{-j!}
<2\,10^{-(N+1)!}=2q_N^{-(N+1)}.
$$
If $\alpha=a/b$ were rational, the distinct fractions $\alpha,s_N$ would have distance at least $1/(bq_N)$, contradicting the tail bound for large $N$. Thus $\alpha$ is irrational. If it were algebraic of fixed degree $d$, apply the <Liouville approximation theorem> to $s_N$. Even if its reduced denominator is smaller than $q_N$, the lower bound is at least $C_\alpha q_N^{-d}$. The displayed upper bound contradicts this as $N\to\infty$. Hence
$$
\boxed{\alpha\text{ is transcendental}\iff\#\{n:\zeta_n=1\}=\infty}.
$$
The coincidence $0!=1!$ changes only the first rational contribution and has no effect on this criterion.
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