[proofplan] Since $[L:K] = n$, the extension $L$ is an $n$-dimensional $K$-vector space. The $n+1$ elements $1, \alpha, \alpha^2, \ldots, \alpha^n$ must be linearly dependent over $K$. A nontrivial dependence relation yields a nonzero polynomial in $K[t]$ vanishing at $\alpha$, so $\alpha$ is algebraic over $K$ with minimal polynomial of degree at most $n$. [/proofplan] [step:Linear dependence from dimension counting] Let $\alpha \in L$. Consider the list of $n+1$ elements \begin{align*} 1,\; \alpha,\; \alpha^2,\; \ldots,\; \alpha^n \end{align*} in the $K$-vector space $L$. Since $\dim_K L = n$, any set of $n+1$ vectors in $L$ is $K$-linearly dependent. Therefore there exist $a_0, a_1, \ldots, a_n \in K$, not all zero, such that \begin{align*} a_0 + a_1\,\alpha + a_2\,\alpha^2 + \cdots + a_n\,\alpha^n = 0. \end{align*} [/step] [step:Constructing the annihilating polynomial] Define $f(t) = a_0 + a_1\,t + a_2\,t^2 + \cdots + a_n\,t^n \in K[t]$. Because the $a_i$ are not all zero, $f$ is a nonzero polynomial, and by construction $f(\alpha) = 0$. Hence $\alpha$ is algebraic over $K$. The minimal polynomial $P_\alpha$ divides $f$ in $K[t]$, so \begin{align*} \deg P_\alpha \;\leq\; \deg f \;\leq\; n \;=\; [L:K]. \end{align*} Since $\alpha \in L$ was arbitrary, every element of $L$ is algebraic over $K$ with minimal polynomial of degree at most $n$. $\blacksquare$ [/step] [guided] The converse of this theorem is false: an algebraic extension need not be finite. Consider the algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$. Every element $\alpha \in \overline{\mathbb{Q}}$ is algebraic over $\mathbb{Q}$ by definition, yet $[\overline{\mathbb{Q}} : \mathbb{Q}]$ is infinite. To see this, note that $\overline{\mathbb{Q}}$ contains $\mathbb{Q}(\sqrt[n]{2})$ for every $n \geq 1$, and $[\mathbb{Q}(\sqrt[n]{2}) : \mathbb{Q}] = n$. Since a finite extension of degree $d$ can only contain subextensions of degree at most $d$, no single finite degree can accommodate all of these, so $[\overline{\mathbb{Q}} : \mathbb{Q}] = \infty$. The correct relationship is: $L/K$ is finite if and only if $L$ is algebraic over $K$ **and** finitely generated as a field extension over $K$. Algebraic alone is not enough. [/guided]