It is normal if we know the ring $k[X_1,...,X_n]$ is Cohen-Macaulay (CM) ring by the definition which is that the ring has unmixed condition.
However, I got stuck when I try to prove that ring is CM ring by this definition: $R$ is a CM ring if the localization of $R$ at $P$, $R_{P}$ is a CM local ring (I suppose that this means $R_{P}$ is an $R$-module whence $\operatorname{depth}R_{P}=\dim R_{P}=\operatorname{ht}P$).
It is very grateful to me if you help me solve it by this definition. Thanks you in advance.
Let me elaborate on my brief comment.
Let $P$ be a prime ideal of height $d$ in $R=k[x_1, \dotsc, x_n]$. Then Noether normalization (as in Eisenbud's Commutative Algebra, Theorem 13.3) in particular tells us that there is a polynomial ring $S=k[t_1, \dotsc, t_n] \subset R$ such that $P \cap S = (t_1, \dotsc, t_d)$.
In particular $P$ contains the regular sequence $t_1, \dotsc, t_d$ of length $d$, thus $R_P$ is Cohen-Macaulay.
The fact that $S \subset R$ is a finite ring extension is not even needed for our application.