[proofplan] We induct on $m$. The base case $m = 1$ follows from the Structure of Simple Extensions: adjoining a single algebraic element yields a finite extension whose degree equals the degree of the minimal polynomial. For the inductive step, we write $K(\alpha_1, \dots, \alpha_{m+1}) = K(\alpha_1, \dots, \alpha_m)(\alpha_{m+1})$ and apply the Tower Law. The inductive hypothesis gives $[K(\alpha_1, \dots, \alpha_m) : K] < \infty$, and we need $[K(\alpha_1, \dots, \alpha_m)(\alpha_{m+1}) : K(\alpha_1, \dots, \alpha_m)] < \infty$, which follows because $\alpha_{m+1}$ is algebraic over $K$ and hence algebraic over the larger field $K(\alpha_1, \dots, \alpha_m)$. [/proofplan] [step:Base case] Suppose $m = 1$, so $L = K(\alpha_1)$ where $\alpha_1$ is algebraic over $K$. Let $f \in K[x]$ be the minimal polynomial of $\alpha_1$ over $K$, with $\deg f = n$. By the Structure of Simple Extensions, the set $\{1, \alpha_1, \alpha_1^2, \dots, \alpha_1^{n-1}\}$ is a basis for $K(\alpha_1)$ as a $K$-vector space. Therefore $[K(\alpha_1) : K] = n < \infty$. [/step] [step:Inductive step] Assume for some $m \geq 1$ that whenever $\beta_1, \dots, \beta_m$ are algebraic over $K$, the extension $K(\beta_1, \dots, \beta_m)/K$ is finite. Now let $\alpha_1, \dots, \alpha_m, \alpha_{m+1} \in L$ all be algebraic over $K$. Set $F = K(\alpha_1, \dots, \alpha_m)$. Then \begin{align*} K(\alpha_1, \dots, \alpha_m, \alpha_{m+1}) = F(\alpha_{m+1}). \end{align*} By the inductive hypothesis, $[F : K] < \infty$. [guided] We need $[F(\alpha_{m+1}) : F] < \infty$. We know $\alpha_{m+1}$ is algebraic over $K$, but we need it to be algebraic over the possibly larger field $F$. Since $K \subseteq F$, the minimal polynomial of $\alpha_{m+1}$ over $K$ is a polynomial in $F[x]$ that still has $\alpha_{m+1}$ as a root. So $\alpha_{m+1}$ is algebraic over $F$. [/guided] Since $\alpha_{m+1}$ is algebraic over $F$, the base case applied to $F(\alpha_{m+1})/F$ gives $[F(\alpha_{m+1}) : F] < \infty$. Now the Tower Law yields \begin{align*} [K(\alpha_1, \dots, \alpha_{m+1}) : K] = [F(\alpha_{m+1}) : F] \cdot [F : K] < \infty. \end{align*} This completes the induction. $\blacksquare$ [/step]