My question is about Newton polygon.
Let $P(x)=\prod _{i=1}^n (x-\alpha_i)$ is a plynomial of degree $n$ in $F[x]$. Suppose $x \in [k-1,k]$ with $1 \leq k \leq n$. Then prove that \begin{equation}NP(x)=\sum_{i=1}^{n}\text{ord}(\alpha_i)+(k-x)\text{ord}(\alpha_k) , \cdots \cdots (1) \end{equation} where $NP$ is the function on the range $[0,n]$ whose graph is the bottom of the Newton polygon $\mathcal{C}_P$ of the polynomial $P(x)$.
I got a partial proof (couldn't complete it myself) as follows which I copied below:
$\underline{\text{Proof}}:$
For each $I$ of $[1,n]$ let $$ \alpha_I=\prod_{i \in I} \alpha_i.$$ So now $P(x)=\sum_{n=0}^{n}p_ix^i$ with $p_0=1$ and $p_k=\sum_{|I|=n-k} \alpha_I$.
For example, $p_n=\prod_{i=1}^n \alpha_i$.
Now let $M_k=\min_{|I|=k} \text{ord}(\alpha_I)=\text{ord}(\alpha_1 \cdots \alpha_k)=\mu_1+\cdots+\mu_k, \ (\mu_i=\text{ord}(\alpha_i)$ with the convention $M_0=0$. The corner of the Newton polygon at the far left is $(0,M_n)$, that far right is $(n,0)$. The bottom of the Newton polygon is the polygon path connecting all the points $(k,M_{n-k})$. Let's call it $\Gamma_p$. This will follow from the following two claims, together with basic facts about convex regions:
$(i)$ the path $\Gamma_p$ lies below the Newton polygon,
$(ii)$ its vertices (i.e., its extremal points) lie on the Newton polygon.
In order to prove $(1)$, it suffices to show that the actual vertices of $\Gamma_p$ are points of the Newton polygon. The segment from $(n-k,M_k)$ to $(n-k+1,M_{k-1})$ has slope $\text{ord}(\alpha_k)$. The vertices of $\Gamma_p$ are therefore the points $(n-k,M_k)$ for which $\text{ord}(\alpha_{k+1})>\text{ord}(\alpha_k)$. But then $\text{ord}(\alpha_{k+1})>\text{ord}(\alpha_j)$ for all $j \leq k$, and $\text{ord}(p_{n-k})=M_k$, so that $(n-k, M_k)$ is a vertex of $\mathcal{C}_p$.
How does the conclusion prove the equality $(1)$ ?
Can some one complete the proof?