Transcendence measure for the canonical Liouville number

Question

Let $\displaystyle\alpha=\sum_{n=0}^{+\infty}\frac1{10^{n!}}$. It is well-known that $\alpha$ is transcendental. I am looking for a transcendental measure for $\alpha$. That is exercise 11.15 of the book of Masser "Auxilliary polynomials in number theory" : Masser book . Here is what I did. Let $\beta\in\overline{\mathbb Q}$ of degree $d$ and naive height $H$. Set $\displaystyle\alpha_n=\sum_{k=0}^n\frac1{10^{k!}}$. One has $$|\alpha-\beta|\ge|\alpha_n-\beta|-|\alpha-\alpha_n|.$$ It is easy to see that $|\alpha-\alpha_n|\le\frac2{10^{n+1}}$. One has \begin{eqnarray*} |\alpha-\beta|&\ge&|\beta-\alpha_n|-\frac2{10^{(n+1)!}}. \end{eqnarray*} Let $\delta$ be a denominator de $\beta$. Obviously $\delta \le H$. Moreover $|\overline{\beta}|\le 1+H$. So $10^{n!}H$ is a denominator of $\beta-\alpha_n$ and $|\overline{\beta-\alpha_n}|\le 1+H+\alpha$. By Liouville inequality, $$|\beta-\alpha_n|\ge e^{(1-d)\ln(1+H+\alpha)-d\ln(10^{n!}H)}.$$ So, $$|\alpha-\beta|\ge\frac{H^{-d}(1+H+\alpha)^{1-d}}{10^{n!}}-\frac2{10^{(n+1)!}}.$$ But I do not know how to bound $\frac{H^{-d}(1+H+\alpha)^{1-d}}{10^{n!}}-\frac2{10^{(n+1)!}}$ to conclude. Any answer is welcome. Thanks in advance