I am reading "Concrete Mathematics" by Donald Knuth, et al..
There is the following proposition in this book without a proof:
$n^{\alpha_1} (\log n)^{\alpha_2} (\log \log n)^{\alpha_3} \prec n^{\beta_1} (\log n)^{\beta_2} (\log \log n)^{\beta_3} \iff (\alpha_1, \alpha_2, \alpha_3) < (\beta_1, \beta_2, \beta_3)$.
Here '$(\alpha_1, \alpha_2, \alpha_3) < (\beta_1, \beta_2, \beta_3)$' means lexicographic order.
Here '$f(n) \prec g(n)$' means $\lim_{n \to \infty} \frac{f(n)}{g(n)} = 0$.
I proved this proposition as follows, but I am not sure that my proof is a proof which the authors expect.
If my proof is not a proof which the authors expect, please show me a proof which the authors expect.
My proof is the following:
Lemma:
Let $\alpha$ be any real number.
Let $\beta$ be a positive real number.
Then,
$\lim_{x \to \infty} \frac{(\log x)^\alpha}{x^\beta}=0$.
Proof of lemma:
Let $m$ be a natural number such that $\alpha + 1 < m$.
Let $x$ be a positive real number.
By Taylor's Theorem, there exists $\theta \in (0, 1)$ such that $$e^x = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^m}{m!}+\frac{e^\theta}{(m+1)!}x^{m+1} > \frac{x^m}{m!}.$$
If $x > 1$, then $x^m > x^{\alpha+1}$, so $$\frac{e^x}{x^\alpha} > \frac{x^m}{m! x^\alpha} > \frac{x}{m!} \to +\infty(x \to +\infty).$$
$\frac{(\log x)^\alpha}{x^\beta}=\frac{(\frac{\beta \log x}{\beta})^\alpha}{e^{\beta \log x}}=\frac{1}{\beta^\alpha} \frac{(\beta \log x)^\alpha}{e^{\beta \log x}} = \frac{1}{\beta^\alpha} \frac{1}{\frac{e^{\beta \log x}}{(\beta \log x)^\alpha}} \to 0 (x \to \infty)$.
Proof of the proposition:
Let $\alpha_1 < \beta_1$.
$$\frac{n^{\alpha_1} (\log n)^{\alpha_2} (\log \log n)^{\alpha_3}}{n^{\beta_1} (\log n)^{\beta_2} (\log \log n)^{\beta_3}} = \frac{(\log n)^{\alpha_2 - \beta_2} (\log \log n)^{\alpha_3 - \beta_3}}{n^{\beta_1 - \alpha_1}}.$$
If $\alpha_3 \leq \beta_3$, then by the above lemma, it is obvious that $\frac{(\log n)^{\alpha_2 - \beta_2} (\log \log n)^{\alpha_3 - \beta_3}}{n^{\beta_1 - \alpha_1}} \to 0$.
So, we assume that $\alpha_3 > \beta_3$.
$$\frac{(\log n)^{\alpha_2 - \beta_2} (\log \log n)^{\alpha_3 - \beta_3}}{n^{\beta_1 - \alpha_1}} < \frac{(\log n)^{\alpha_2 - \beta_2} (\log n)^{\alpha_3 - \beta_3}}{n^{\beta_1 - \alpha_1}} = \frac{(\log n)^{\alpha_2 - \beta_2+\alpha_3 - \beta_3}}{n^{\beta_1 - \alpha_1}} \to 0$$ by the above lemma.
Let $\alpha_1 = \beta_1$ and $\alpha_2 < \beta_2$.
$$\frac{n^{\alpha_1} (\log n)^{\alpha_2} (\log \log n)^{\alpha_3}}{n^{\beta_1} (\log n)^{\beta_2} (\log \log n)^{\beta_3}} = \frac{(\log \log n)^{\alpha_3 - \beta_3}}{(\log n)^{\beta_2 - \alpha_2}} \to 0$$ by the above lemma.
Let $\alpha_1 = \beta_1$ and $\alpha_2 = \beta_2$ and $\alpha_3 < \beta_3$.
$$\frac{n^{\alpha_1} (\log n)^{\alpha_2} (\log \log n)^{\alpha_3}}{n^{\beta_1} (\log n)^{\beta_2} (\log \log n)^{\beta_3}} = \frac{1}{(\log \log n)^{\beta_3 - \alpha_3}} \to 0.$$