(This question is related to some of my most recent questions, so maybe the question setup might be familiar to one or the other. ;) )
I'm in need of some help / hints / notes about a specific setup as follows. I know it is a longer question, but maybe someone has the patience and can help me:
Let $X = \text{Spec}(A)$ be a smooth variety (integral separated $K$-scheme of finite type) with generic point $y_0$ over some algebraically closed field $K$. We furthermore suppose we have a regular sequence $f_1, \dots, f_p \in A$, such that $Y_p = V(f_1, \dots, f_p)$ is a nonempty smooth closed subvariety with generic point $y_p$, i.e. $\overline{\{y_p\}} = Y_p$. In particular we can assume that the ideal $(f_1, \dots, f_p) \subseteq A$ is prime. We always consider closure of points with their integral subscheme structure.
Now as $Y_p$ is smooth, also $V(f_1, \dots, f_i)$ is smooth along $Y_p$ for any $1 \leq i \leq p$. This yields for any $i = 1, \dots, p$, that $Y_p$ is contained in exactly one irreducible component $Y_i = \overline{\{y_i\}}$ of $V(f_1, \dots, f_i)$ (because otherwise $\mathcal O_{V(f_1, \dots, f_i), y_p}$ would not be an integral domain, in contradiction to it being regular and local).
We thus get an inclusion of irreducible components $Y_i$ of $V(f_1, \dots, f_i)$ by $$Y_p := Z \subsetneq Y_{p-1} \subsetneq \cdots \subsetneq Y_1 \subsetneq Y_0 := X = \overline{\{y_0\}},$$ where $Y_{i+1}$ is of codimension $1$ in $Y_i$.
Hence if we consider any $f_{i+1}$ as an element in the function field $K(Y_{i}) = \mathcal O_{Y_i, y_i}$, then $f_{i+1} \in K(Y_{i})^\times$ (otherwise $Y_{i+1}$ would not be of codimension $1$). Note that $K(Y_i) = \text{Frac}(\mathcal O_{Y_i, y_{i+1}})$ and we have the order of vanishing
$$ \text{ord}_{Y_{i+1}}(f_{i+1}). $$
In a proof of something else I read it was used that $$ \prod_{i = 0}^{p-1} \text{ord}_{Y_{i+1}}(f_{i+1}) = 1, $$ however I do not see the equality. My professor said the equality should not be that surprising, but could not give an argument on the fly.
My thoughts so far: The question kind of boils down to showing that each factor in the product is $1$. Let's take for example the first case, i.e. $\text{ord}_{Y_1}(f_1) = 1$. Here we are in the convenient situation that $\mathcal O_{X, y_1}$ is regular, hence a DVR. But if $\text{ord}_{Y_1}(f_1) = 1$, then this would imply that $f_1$ is a uniformizer in the maximal ideal of $\mathcal O_{X,y_1}$. If $\mathfrak p \in \text{Spec}(A)$ corresponds to $y_1$, then it is a minimal prime over $f_1$ with $(f_1) \subseteq \mathfrak p \subseteq (f_1, \dots, f_p)$. But I see no reason why $f_1$ should be a generator of the maximal ideal $\mathfrak p A_{\mathfrak p}$.
A great thanks to anyone who has read until here!
With the help of Mohan's comment, I would have proved the statement in the following way:
We consider the regular local domain $\mathcal O_{X, \mathfrak p} = A_\mathfrak p$, where $\mathfrak p = (f_1, \dots, f_p) \in \text{Spec}(A)$. Then $f_1, \dots, f_p$ form a coordinate system for $A_\mathfrak p$, in particular the sequence stays regular. Consider now the ideal $\mathfrak q := (f_1, \dots, f_i) \subseteq A_\mathfrak p$. Then, as $f_1, \dots, f_p$ are a coordinate system for $A_\mathfrak p$, the local ring $A_\mathfrak p/\mathfrak q$ is regular (hence a domain) of dimension $p-i$. Note that the prime ideal $\mathfrak q'$ in $A$ corresponding to $\mathfrak q = \mathfrak q'A_\mathfrak p$ is clearly the unique minimal prime over $(f_1, \dots, f_i)$ satisfying $(f_1, \dots, f_i) \subseteq \mathfrak q' \subseteq \mathfrak p$.
So we now consider the irreducible component $Y_i$ and use the notation as above, in particular we assume that $Y_i = \text{Spec}(A/\mathfrak q')$. Consider minimal prime $\mathfrak a$ over $f_{i+1} + \mathfrak q' \in A/\mathfrak q'$ determining $Y_{i+1}$, with corresponding prime ideal $\mathfrak a' \in \text{Spec}(A)$ satisfying $(f_{i+1}, \mathfrak q') \subseteq \mathfrak a' \subseteq \mathfrak p$. We then have (with some abuse of notation) $$ \mathcal O_{Y_i, y_{i+1}} \cong (A/\mathfrak q')_\mathfrak a \cong (A_\mathfrak p)_{\mathfrak a'} / \mathfrak q' (A_\mathfrak p)_{\mathfrak a'} \cong (A_\mathfrak p/\mathfrak q)_{\mathfrak a'}, $$
which is a regular local domain as localization of a regular local domain. By the argument above, the maximal ideal $\mathfrak a'(A_\mathfrak p/\mathfrak q)_{\mathfrak a'} $ is generated by $f_{i+1} + \mathfrak q$, so we get $$ \text{ord}_{Y_{i+1}}(f_{i+1}) = 1. $$