[proofplan] We compute the degree by intersecting $C$ with a hyperplane. Pulling a hyperplane equation on $\mathbb P^n_k$ back along the degree-$n$ Veronese map gives a nonzero homogeneous binary form of degree $n$. Such a form has a zero divisor of degree $n$ on $\mathbb P^1_k$, and the Veronese map is a closed immersion onto $C$, so this divisor is exactly the hyperplane section of $C$ with multiplicities preserved. [/proofplan] [step:Pull back a general hyperplane to a binary form of degree $n$] Let $S:=k[X_0,\dots,X_n]$ be the homogeneous coordinate ring of $\mathbb P^n_k$, and let $R:=k[s,t]$ be the homogeneous coordinate ring of $\mathbb P^1_k$. Let $H\subset \mathbb P^n_k$ be a hyperplane with defining linear form \begin{align*} L:=a_0X_0+a_1X_1+\cdots+a_nX_n\in S_1. \end{align*} where $(a_0,\dots,a_n)\in k^{n+1}\setminus\{0\}$. The morphism $\nu_n$ induces the homogeneous coordinate homomorphism \begin{align*} \nu_n^\#: S \to R,\qquad X_i \mapsto s^{n-i}t^i \quad \text{for } 0\le i\le n. \end{align*} Therefore the pullback of $L$ is the homogeneous binary form \begin{align*} F:=\nu_n^\#(L)=a_0s^n+a_1s^{n-1}t+\cdots+a_nt^n\in R_n. \end{align*} Since the monomials $s^n,s^{n-1}t,\dots,t^n$ form a $k$-basis of $R_n$ and not all $a_i$ are zero, we have $F\ne 0$. Define $V_+(F)\subset \mathbb P^1_k$ to be the projective zero scheme cut out by the homogeneous form $F$, equivalently the effective Cartier divisor associated to the principal homogeneous equation $F=0$. Hence no hyperplane $H$ contains $C$, and the scheme-theoretic pullback of $H\cap C$ to $\mathbb P^1_k$ is the effective divisor $V_+(F)$. [/step] [step:Count the zero divisor of the pulled-back form on $\mathbb P^1_k$] Because $F\in k[s,t]$ is a nonzero homogeneous form of degree $n$ and $k$ is algebraically closed, its zero divisor on $\mathbb P^1_k$ has degree $n$, counted with multiplicity. Indeed, if $t\neq 0$, write $u:=s/t$ on the affine chart $D_+(t)\cong \mathbb A^1_k$, where $\mathbb A^1_k:=\operatorname{Spec} k[u]$ denotes the affine line over $k$. On this chart, \begin{align*} F(s,t)=t^n f(u), \end{align*} where \begin{align*} f(u):=a_0u^n+a_1u^{n-1}+\cdots+a_n\in k[u]. \end{align*} If $f$ is not the zero polynomial, then by the factorisation theorem for polynomials over an [algebraically closed field](/page/Algebraically%20Closed%20Field), $f$ has degree $\deg f$ and exactly $\deg f$ zeros in $\mathbb A^1_k$ counted with multiplicity. If $\deg f<n$, the remaining multiplicity $n-\deg f$ occurs at the point $[1:0]\in \mathbb P^1_k$. Thus in all cases the divisor $V_+(F)$ has total degree $n$. This is the standard homogeneous form version of the [fundamental theorem of algebra](/theorems/347) over an algebraically closed field, with the multiplicity at $[1:0]$ accounting for the possible drop in the affine degree of $f$. [guided] The purpose of this step is to translate the geometric intersection problem into the elementary fact that a degree-$n$ homogeneous binary form has $n$ projective zeros when multiplicities are counted. We work on the affine chart $D_+(t)\subset \mathbb P^1_k$, where $t\neq 0$. This chart is isomorphic to the affine line $\mathbb A^1_k:=\operatorname{Spec} k[u]$. Define the affine coordinate \begin{align*} u:=s/t. \end{align*} Then every point of this chart has the form $[u:1]$, and evaluating $F$ gives \begin{align*} F(u,1)=a_0u^n+a_1u^{n-1}+\cdots+a_n. \end{align*} Define the polynomial \begin{align*} f:k&\to k \end{align*} \begin{align*} u&\mapsto a_0u^n+a_1u^{n-1}+\cdots+a_n. \end{align*} Equivalently, as an element of $k[u]$, write \begin{align*} f(u):=a_0u^n+a_1u^{n-1}+\cdots+a_n. \end{align*} If $f$ has degree $d$, then the algebraic closedness of $k$ implies that $f$ factors as \begin{align*} f(u)=c\prod_{j=1}^r (u-\lambda_j)^{m_j}, \end{align*} where $c\in k^\times$, the points $\lambda_1,\dots,\lambda_r\in k$ are distinct, the integers $m_1,\dots,m_r$ are positive, and \begin{align*} m_1+\cdots+m_r=d. \end{align*} Thus the zeros in the chart $D_+(t)$ contribute total multiplicity $d$. If $d=n$, there is no missing multiplicity. If $d<n$, then the coefficient of $u^n$ and perhaps further leading coefficients vanish, and the missing multiplicity occurs at the point $[1:0]$. To see this, use the other affine chart $D_+(s)$ with coordinate \begin{align*} v:=t/s. \end{align*} On this chart, \begin{align*} F(1,v)=a_0+a_1v+\cdots+a_nv^n. \end{align*} The order of vanishing at $v=0$, which is the point $[1:0]$, is exactly $n-d$. Hence the total multiplicity of the zero divisor of $F$ on $\mathbb P^1_k$ is \begin{align*} d+(n-d)=n. \end{align*} Therefore $V_+(F)$ is an effective divisor of degree $n$ on $\mathbb P^1_k$. [/guided] [/step] [step:Transfer the divisor count from $\mathbb P^1_k$ to the rational normal curve] By the theorem that the degree-$n$ Veronese map $\nu_n:\mathbb P^1_k\to \mathbb P^n_k$ is a closed immersion onto its image $C$, we identify $\mathbb P^1_k$ isomorphically with $C$. Under this isomorphism, the hyperplane section $H\cap C$ pulls back to the divisor $V_+(F)$ on $\mathbb P^1_k$, because $F=\nu_n^\#(L)$ is the pullback of the defining equation $L$ of $H$. Isomorphisms preserve local rings and hence preserve intersection multiplicities of Cartier divisors. Therefore \begin{align*} \deg(H\cap C)=\deg V_+(F)=n. \end{align*} [/step] [step:Conclude that the degree of $C$ is $n$] By the definition of the degree of a projective curve, $\deg C$ is the degree of the hyperplane section $H\cap C$ for a general hyperplane $H\subset \mathbb P^n_k$ not containing $C$. The preceding steps show that every hyperplane $H$ with nonzero defining linear form has $H\not\supset C$ and satisfies \begin{align*} \deg(H\cap C)=n. \end{align*} Consequently, \begin{align*} \deg C=n. \end{align*} This proves the theorem. [/step]