Fibre cardinality of an unramified morphism Let $\varphi: X \to Y$ be a finite, dominant, unramified morphism of varieties over an algebraically closed field. If necessary, we can assume $X$ and $Y$ to be nonsingular. I am trying to prove that 
$$\mathrm{deg}(\varphi):=[K(Y):K(X)]=|\varphi^{-1}(P)|$$
for every point $P\in Y$. The statement is very easy to prove for curves. However, I am completely stuck trying to prove it for higher dimensions. I cannot find this statement or a similar one in literature, it would also be a great help if someone could point me there.
 A: Claim Let $\phi:X\to Y$ be a finite étale morphism (i.e., flat and unramified) of reduced schemes of finite type over an algebraically closed field. Assume that $Y$ is irreducible and Cohen-Macaulay and $\phi$ is dominant on every irreducible component of $X$. Then for a closed point $P\in Y$, the number of pre-images of $P$, denoted by $|\phi^{-1}(P)|$ is independent of $P$. Define $\deg\phi$ to be this value. 
If $X$ and $Y$ are both irreducible, this value is equal to $[K(Y):K(X)]$.
Proof Since $Y$ is connected, the statement is local and we may assume that $Y$ is affine and hence quasi-projective. Let $P\in Y$ and if $\dim Y>1$, then let $H\subseteq Y$ be a general effective very ample divisor such that $P\in H$. By the assumptions $H$ is again an irreducible reduced Cohen-Macaulay scheme of finite type over an algebraically closed  field. Replace $Y$ with $H$ and $X$ with $\phi^{-1}H$. Notice that the original assumptions remain true, in particular $\phi^{-1}H\to H$ is étale and for any $P,Q\in Y$ and any two general effective very ample divisors $H_P,H_Q\in Y$ such that $P\in H_P$ and $Q\in H_Q$ it follows that $H_P\cap H_Q\neq\emptyset$. 
Therefore we may assume that $\dim Y=1$. 
It follows that $\dim X=1$ and since $\phi$ is étale, the irreducible components of $X$ are disjoint, so we may assume that $X$ is also irreducible. Let $\widetilde Y\to Y$ be a resolution of singularities of $Y$ and consider the base change $\widetilde\phi: \widetilde X\to \widetilde Y$. Since $\phi$ is étale, so is $\widetilde\phi$ and hence $\widetilde X$ is also non-singular. In other words we may assume that $X$ and $Y$ are nonsingular curves.
In that case $\{P\}\subset Y$ is a divisor and it is well-known that $\deg\phi^*P=[K(Y):K(X)]$. 
We obtain that $\deg\phi^*(P)$ is independent of $P\in Y$ for arbitrary points and since $\phi$ is étale, this implies that if $P\in Y$ is a closed point, then $|\phi^{-1}(P)|=\deg\phi^*(P)$ is independent of $P$.
Now if $X$ and $Y$ are both irreducible at the start, then since the value $|\phi^{-1}(P)|$ is independent of $P$ for closed points, it is enough to check that value at one particular point. As $\phi$ is étale, the field extension $K(Y)\subseteq K(X)$ is (finite) separable and hence may be generated by a single element with a minimal polynomial of degree $[K(Y):K(X)]$. This shows that $Y$ may be embedded in some projective space $\mathbb P^N$ such that $X$ is birational to a hypersurface in $\mathbb P^N\times \mathbb A^1$ such that (as a rational map) $\phi$ is the composition of the projection to $\mathbb P^N$ and the birational map on $X$. It follows that for a general (closed) point the equality $|\phi^{-1}(P)|=[K(Y):K(X)]$ holds.
$\square$
Comments
1) Perhaps someone has a simpler argument for the last paragraph, but I don't see how one can easily relate the degree of the field extension to the number of points in a fiber. The appearance of this field extension degree in the proof of the first statement would need further work, because it is the degree of the field extension of the complete intersection curves we obtain by taking hyperplane cuts. Maybe it is obvious, but I am not sure how to prove easily that the degree of that field extension is the same as the degree of the original. One way to prove it is a variation of the last paragraph above.
2) Obviously the statement is only for closed points, but I am sure that's what Jesko meant.
3) Just for the record: the above statement implies the one in the question in case $X$ is Cohen-Macaulay and $Y$ is non-singular as those together imply that $\phi$ is flat.
4) See alternative proof in the comments above by Damian Rössler.
A: After I wrote the comments above, I found the following reference : 
Formula (12.6.2), p. 329 in Görtz-Wedhorn, Algebraic Geometry I, Viehweg & Teubner Verlag
for (a generalisation of) the equality you are looking for, when $\phi$ is assumed flat (which is true if you assume that $X$ and $Y$ are non-singular, as pointed out in the comments of K. M. Pera and S. Kovacs). 
