Some doubts on a classical example of Liouville Number

Question

Theorem (Liouville) Given a real algebraic number $\alpha$ of degree $>1$, there is a positive constant $c=c(\alpha)$ s.t. for all rational numbers $p/q$ with $(p,q)=1,q>0$, we have $$\lvert\alpha-p/q\rvert>\frac{c(\alpha)}{q^n}.$$

The author claims that, using this theorem, one can prove that $\sum_{n=0}^\infty\frac{1}{10^{n!}}$ is transcendental, and he argues as follows. Suppose not and call the sum $\alpha$. Consider the partial sums $\frac{p_k}{q_k}=\sum_{n=0}^k\frac{1}{10^{n!}}.$ Clearly $\lvert\alpha- \frac{p_k}{q_k}\rvert<\frac{c}{10^{(k+1)!}}$ for some constant $c>0$.

If $\alpha$ were algebraic of degree $m$, then by Liouville's Theorem, the left-hand side would be greater than $c(\alpha)/10^{k!m}$ and for $k$ sufficiently large, this is a contradiction.

Questions: (1) when he claims "If $\alpha$ were algebraic of degree $m$", I guess he deals with $m\ge2$, so that he can apply Liouville's result. But what about the case $m=1$? In other words, why is $\alpha$ not rational?

(2) the proof ends with the phrase "for $k$ sufficiently large, this is a contradiction". Unfortunately, I don't where the contradiction is.

Any help would be appreciated.

Answer 1

To see that $\alpha$ must be irrational, remember that rational numbers always have eventually periodic decimal (or indeed any base) expansions. But $\alpha$ clearly doesn't.

For question $2$, compare the denominators: for any $m$, when $k$ is sufficiently large we have $k!\cdot m<(k+1)!$, giving in turn $${1\over k!\cdot m}>{1\over (k+1)!}.$$ Now first imagine $c=c(\alpha)$, and then note that even if $c\not=c(\alpha)$ we still eventually get a contradiction as $k$ increases.