$\text{Newton Polygons for Polynomials}$
There is a lemma in the book $ \ \text{p-adic numbers, p-adic analysis and zet-functions} $ of the author $ \text{Neal Koblitz} \ $ which I mentioned below:
$ \text{Lemma 4}: \ $ Let $ f(X)=(1-\frac{X}{\alpha_1})(1-\frac{X}{\alpha_2}) \cdots (1-\frac{X}{\alpha_n})$ be the factorisation of $ f(X)$ in terms of its roots $ \alpha_i \in \Omega$. Let $ \lambda_i= \text{ord}_p (\frac{1}{\alpha_i})$. Then, if $ \lambda $ is a slope of the Newton polygon having length $l$, it follows that precisely $ l$ of the $ \lambda_i$ are equal to $ \lambda$.
($\Omega$ is a field which is the completion of the algebraic closure of the p-adic field $ \mathbb{Q}_p$)
In other words, the slopes of the Newton polygon of $f(X)$ are (counting multiplicity) the p-adic ordinals of the reciprocal roots of $f(X)$.
This is all about the Lemma.
My question-
Is the length $l$ of the Newton Polygon a natural number?
If it is true, then why?
According to the first statement of the Lemma, it says $ \text{precisely $l$ of the $ \lambda_i$ are equal to $ \lambda$}$, which imply $ l$ must be a natural number.
Though I could not explain it why it is a natural number or why should the length of the newton polygon be a natural number.
Please someone explain it.