[proofplan] We construct the normal closure explicitly. Since $K/k$ is finite, write $K = k(\alpha_1, \ldots, \alpha_r)$ for finitely many algebraic elements. The normal closure $\tilde{K}$ is the splitting field of the product $f = \prod_{i=1}^{r} \operatorname{min}(\alpha_i, k)$ over $k$. To bound $[\tilde{K} : k]$, we use the primitive element theorem to reduce to a single generator $K = k(\alpha)$, show $\tilde{K}$ is the splitting field of $\operatorname{min}(\alpha, k)$, and bound the degree by $n!$ using the tower law and degree bounds for adjunction of roots. [/proofplan]

[step:Reduce to a single generator and identify the normal closure as a splitting field] Since $K/k$ is finite, it is finitely generated as a field extension. Write $K = k(\alpha_1, \ldots, \alpha_r)$ for algebraic elements $\alpha_i \in K$. The normal closure $\tilde{K}$ of $K/k$ inside $\overline{k}$ is the smallest normal extension of $k$ containing $K$. Equivalently, $\tilde{K}$ is the splitting field over $k$ of the polynomial \begin{align*} f = \prod_{i=1}^{r} \operatorname{min}(\alpha_i, k) \in k[x]. \end{align*} This is because a normal extension of $k$ containing $K$ must contain all roots of each $\operatorname{min}(\alpha_i, k)$ (by the definition of normality: every irreducible polynomial in $k[x]$ that has one root in the extension splits completely), and the splitting field of $f$ is the smallest such extension. Since $f$ has degree $\sum_{i=1}^{r} \deg \operatorname{min}(\alpha_i, k) \le \sum_{i=1}^{r} [K : k] = rn$ (a finite integer), the splitting field $\tilde{K}$ is a finite extension of $k$ by the [Existence and Uniqueness of Splitting Fields](/theorems/1258). In particular, $[\tilde{K} : k] < \infty$. [/step]

[step:Bound $[\tilde{K} : k]$ by $n!$ using successive root adjunctions] To obtain the sharper bound $[\tilde{K} : k] \mid n!$, we may assume (by using the primitive element theorem if $K/k$ is separable, or by direct argument otherwise) that we work with the minimal polynomials of the generators. Let $f_i = \operatorname{min}(\alpha_i, k)$ with $d_i = \deg f_i = [k(\alpha_i) : k]$. Since $k(\alpha_i) \subset K$ and $[K : k] = n$, the tower law gives $d_i \mid n$, hence $d_i \le n$. We construct $\tilde{K}$ by successively adjoining all roots of $f_1, \ldots, f_r$. Consider first $f_1$. Let $\beta_1 = \alpha_1, \beta_2, \ldots, \beta_{d_1}$ be the roots of $f_1$ in $\overline{k}$. Set $L_0 = k$ and define $L_j = L_{j-1}(\beta_j)$ for $j = 1, \ldots, d_1$. At each stage, $\beta_j$ is a root of $f_1 \in k[x] \subset L_{j-1}[x]$, so $[L_j : L_{j-1}] \le \deg f_1 = d_1$ (the minimal polynomial of $\beta_j$ over $L_{j-1}$ divides $f_1$). Moreover, each successive adjunction adds a root that may already be in $L_{j-1}$, and the degree $[L_j : L_{j-1}]$ divides $d_1$ and strictly decreases with each new root (it divides $d_1 - (j-1)$ since $f_1$ has at least $j-1$ roots in $L_{j-1}$). More precisely, after adjoining $j-1$ roots, $f_1$ has a factor of degree at least $j-1$ over $L_{j-1}$, so $[L_j : L_{j-1}] \le d_1 - (j-1)$. Repeating this for all polynomials $f_1, \ldots, f_r$, the total degree satisfies: \begin{align*} [\tilde{K} : k] &\le \prod_{i=1}^{r} d_i! \end{align*} We now give the clean bound. Each $k$-embedding $\sigma: K \to \overline{k}$ is determined by the images $\sigma(\alpha_1), \ldots, \sigma(\alpha_r)$, and there are at most $n$ such embeddings (since $[K : k] = n$ and the number of $k$-embeddings of $K$ into $\overline{k}$ is at most $[K : k]$). The normal closure $\tilde{K}$ is generated over $k$ by $\bigcup_{\sigma} \sigma(K)$ where $\sigma$ ranges over all $k$-embeddings $K \to \overline{k}$. Equivalently, $\tilde{K}$ is the compositum of $\sigma_1(K), \ldots, \sigma_m(K)$ where $\sigma_1, \ldots, \sigma_m$ are the distinct $k$-embeddings of $K$ into $\overline{k}$, with $m \le n$. Each $\sigma_j(K)$ has degree $n$ over $k$ (since $\sigma_j$ is an isomorphism $K \xrightarrow{\sim} \sigma_j(K)$). By the tower law applied to the compositum, each successive adjunction satisfies $[\tilde{K}_j : \tilde{K}_{j-1}] \le n$ where $\tilde{K}_j = \tilde{K}_{j-1} \cdot \sigma_j(K)$, and in fact $[\tilde{K}_j : \tilde{K}_{j-1}]$ divides $[\sigma_j(K) : k] = n$. Since $\tilde{K}_0 = K = \sigma_1(K)$ and we adjoin at most $m - 1 \le n - 1$ further conjugates: \begin{align*} [\tilde{K} : k] \;\Big|\; n \cdot (n-1) \cdot (n-2) \cdots 1 = n!. \end{align*} More precisely, $[\tilde{K} : k]$ divides $n!$ because $\operatorname{Gal}(\tilde{K}/k)$ (when $\tilde{K}/k$ is separable) embeds into $S_n$ via its action on the $n$ roots of $\operatorname{min}(\alpha, k)$ (in the case $K = k(\alpha)$), and a subgroup of $S_n$ has order dividing $n!$.[/step]

[guided]To get the bound $[\tilde{K} : k] \mid n!$, we use the Galois-theoretic structure. First, observe that $\tilde{K}$ is the splitting field of the polynomial $f = \prod_{i=1}^{r} \operatorname{min}(\alpha_i, k)$ over $k$. In the special case where $K = k(\alpha)$ for a single element (which we can always achieve over separable extensions by the [Primitive Element Theorem](/theorems/1267)), $\tilde{K}$ is the splitting field of $g = \operatorname{min}(\alpha, k)$, which has degree $n = [K : k]$. Let $\alpha = \alpha_1, \alpha_2, \ldots, \alpha_n$ be the roots of $g$ in $\overline{k}$ (counted with multiplicity; there are exactly $n$ since $\deg g = n$). Then $\tilde{K} = k(\alpha_1, \ldots, \alpha_n)$. We build $\tilde{K}$ by adjoining roots one at a time. Set $L_0 = k$ and $L_j = L_{j-1}(\alpha_j)$ for $j = 1, \ldots, n$. At step $j$, the element $\alpha_j$ satisfies $g(\alpha_j) = 0$, and $g$ already has at least $j - 1$ roots in $L_{j-1}$ (namely $\alpha_1, \ldots, \alpha_{j-1}$). So in $L_{j-1}[x]$, the polynomial $g$ factors as $(x - \alpha_1) \cdots (x - \alpha_{j-1}) \cdot h_j(x)$ where $h_j \in L_{j-1}[x]$ has $\deg h_j = n - (j-1)$. Since $\alpha_j$ is a root of $h_j$, the minimal polynomial of $\alpha_j$ over $L_{j-1}$ divides $h_j$, so: \begin{align*} [L_j : L_{j-1}] \le \deg h_j = n - (j - 1). \end{align*} Multiplying over all steps: \begin{align*} [\tilde{K} : k] = [L_n : k] = \prod_{j=1}^{n} [L_j : L_{j-1}] \le n \cdot (n-1) \cdot (n-2) \cdots 1 = n!. \end{align*} Moreover, at each step $[L_j : L_{j-1}]$ divides $\deg h_j = n - (j-1)$ (since the minimal polynomial divides $h_j$ and $h_j$ is monic), and so $[\tilde{K} : k]$ divides $n!$. As an alternative perspective: any automorphism $\sigma \in \operatorname{Aut}_k(\tilde{K})$ permutes the $n$ roots $\alpha_1, \ldots, \alpha_n$ of $g$, and $\sigma$ is determined by this permutation (since $\tilde{K} = k(\alpha_1, \ldots, \alpha_n)$). This gives an injective homomorphism $\operatorname{Aut}_k(\tilde{K}) \hookrightarrow S_n$. If $\tilde{K}/k$ is separable (which holds when $K/k$ is separable), then $\tilde{K}/k$ is Galois (splitting field of a separable polynomial), so $|\operatorname{Gal}(\tilde{K}/k)| = [\tilde{K} : k]$ by the [Order of the Galois Group](/theorems/3325). Since $\operatorname{Gal}(\tilde{K}/k)$ embeds into $S_n$, its order divides $|S_n| = n!$ by Lagrange's theorem. Hence $[\tilde{K} : k]$ divides $n!$.[/guided]