[proofplan] We prove each direction separately. For the forward direction ($d \mid n$ implies $\mathbb{F}_{p^d} \subset \mathbb{F}_{p^n}$), we show that $p^d - 1 \mid p^n - 1$ by a polynomial divisibility argument, which forces $t^{p^d} - t \mid t^{p^n} - t$ in $\mathbb{F}_p[t]$, so every root of $t^{p^d} - t$ (i.e., every element of $\mathbb{F}_{p^d}$) is also a root of $t^{p^n} - t$ (i.e., an element of $\mathbb{F}_{p^n}$). For the converse ($\mathbb{F}_{p^d} \subset \mathbb{F}_{p^n}$ implies $d \mid n$), we observe that $\mathbb{F}_{p^n}$ is a vector space over $\mathbb{F}_{p^d}$, so $p^n$ is a power of $p^d$, which forces $d \mid n$. [/proofplan] [step:Show that $d \mid n$ implies $p^d - 1 \mid p^n - 1$] Assume $d \mid n$, so $n = dm$ for some positive integer $m$. Consider the polynomial $x^m - 1 \in \mathbb{Z}[x]$. We have the factorisation \begin{align*} x^m - 1 = (x - 1)(x^{m-1} + x^{m-2} + \cdots + x + 1). \end{align*} In particular, $(x - 1) \mid (x^m - 1)$ in $\mathbb{Z}[x]$. Substituting $x = p^d$: \begin{align*} (p^d - 1) \mid (p^{dm} - 1) = (p^n - 1). \end{align*} Explicitly, $p^n - 1 = (p^d - 1)\bigl((p^d)^{m-1} + (p^d)^{m-2} + \cdots + p^d + 1\bigr)$, so the quotient is an integer. [guided] The goal is to establish a divisibility relation between $p^d - 1$ and $p^n - 1$. This is a purely number-theoretic step that does not yet involve fields. Since $d \mid n$, write $n = dm$ for some $m \in \mathbb{N}$. The algebraic identity \begin{align*} x^m - 1 = (x - 1)(x^{m-1} + x^{m-2} + \cdots + x + 1) \end{align*} holds in $\mathbb{Z}[x]$ (verify by expanding the right-hand side: each term $x^k$ for $1 \le k \le m-1$ appears once with coefficient $+1$ from $x \cdot x^{k-1}$ and once with coefficient $-1$ from $-1 \cdot x^k$, so all intermediate terms cancel, leaving $x^m - 1$). Evaluating at $x = p^d \in \mathbb{Z}$ gives \begin{align*} (p^d)^m - 1 = (p^d - 1)\bigl((p^d)^{m-1} + (p^d)^{m-2} + \cdots + p^d + 1\bigr). \end{align*} Since $(p^d)^m = p^{dm} = p^n$, this reads $p^n - 1 = (p^d - 1) \cdot q$ where $q = \sum_{j=0}^{m-1} (p^d)^j$ is a positive integer. Therefore $p^d - 1 \mid p^n - 1$. Why do we care about this divisibility? The finite field $\mathbb{F}_{p^k}$ is the splitting field of $t^{p^k} - t$ over $\mathbb{F}_p$, and its elements are exactly the roots of this polynomial. The exponents $p^k - 1$ control the multiplicative structure: $\mathbb{F}_{p^k}^\times$ is cyclic of order $p^k - 1$. The divisibility $p^d - 1 \mid p^n - 1$ will let us show that $t^{p^d} - t \mid t^{p^n} - t$, which is the polynomial containment that produces the subfield inclusion. [/guided] [/step] [step:Deduce that $t^{p^d} - t \mid t^{p^n} - t$ in $\mathbb{F}_p[t]$] From the previous step, $p^d - 1 \mid p^n - 1$. Write $p^n - 1 = (p^d - 1) \cdot q$ for a positive integer $q$. Every element $\alpha \in \mathbb{F}_{p^d}$ satisfies $\alpha^{p^d} = \alpha$ (since $\mathbb{F}_{p^d}$ is the set of roots of $t^{p^d} - t$ in an algebraic closure of $\mathbb{F}_p$). We show that every such $\alpha$ also satisfies $\alpha^{p^n} = \alpha$. If $\alpha = 0$, then $\alpha^{p^n} = 0 = \alpha$. If $\alpha \ne 0$, then $\alpha \in \mathbb{F}_{p^d}^\times$, so $\alpha^{p^d - 1} = 1$ (since $\mathbb{F}_{p^d}^\times$ is a group of order $p^d - 1$). Therefore \begin{align*} \alpha^{p^n - 1} = \alpha^{(p^d - 1) \cdot q} = \bigl(\alpha^{p^d - 1}\bigr)^q = 1^q = 1. \end{align*} Multiplying both sides by $\alpha$ gives $\alpha^{p^n} = \alpha$. Since every root of $t^{p^d} - t$ is a root of $t^{p^n} - t$, and $t^{p^d} - t$ has no repeated roots (its formal derivative is $p^d t^{p^d - 1} - 1 = -1 \ne 0$ in $\mathbb{F}_p[t]$, so $\gcd(t^{p^d} - t, \, (t^{p^d} - t)') = \gcd(t^{p^d} - t, -1) = 1$), the $p^d$ distinct roots of $t^{p^d} - t$ are all roots of $t^{p^n} - t$. Therefore $t^{p^d} - t \mid t^{p^n} - t$ in $\mathbb{F}_p[t]$. [guided] We must bridge the gap between the integer divisibility $p^d - 1 \mid p^n - 1$ and the polynomial divisibility $t^{p^d} - t \mid t^{p^n} - t$ in $\mathbb{F}_p[t]$. The argument goes through the roots. Fix an algebraic closure $\overline{\mathbb{F}_p}$ of $\mathbb{F}_p$. The finite field $\mathbb{F}_{p^d}$ is, by construction, the splitting field of $t^{p^d} - t$ over $\mathbb{F}_p$, and its $p^d$ elements are exactly the roots of $t^{p^d} - t$ in $\overline{\mathbb{F}_p}$. (The polynomial $t^{p^d} - t$ has degree $p^d$ and no repeated roots, as we verify below, so it has exactly $p^d$ distinct roots.) We check the separability claim. The formal derivative of $t^{p^d} - t$ in $\mathbb{F}_p[t]$ is \begin{align*} \frac{d}{dt}(t^{p^d} - t) = p^d \cdot t^{p^d - 1} - 1 = 0 \cdot t^{p^d - 1} - 1 = -1, \end{align*} where $p^d \cdot t^{p^d - 1} = 0$ because $p^d \equiv 0 \pmod{p}$ in $\mathbb{F}_p$. Since the derivative is the nonzero constant $-1$, we have $\gcd(t^{p^d} - t, -1) = 1$, confirming that $t^{p^d} - t$ has no repeated roots. Now take any $\alpha \in \mathbb{F}_{p^d}$, so $\alpha^{p^d} = \alpha$. We want to show $\alpha^{p^n} = \alpha$. The case $\alpha = 0$ is immediate: $0^{p^n} = 0$. For $\alpha \ne 0$: the element $\alpha$ lies in $\mathbb{F}_{p^d}^\times$, a multiplicative group of order $p^d - 1$. By Lagrange's theorem applied to this finite group, $\alpha^{p^d - 1} = 1$. Using $p^n - 1 = (p^d - 1) \cdot q$: \begin{align*} \alpha^{p^n - 1} = \alpha^{(p^d - 1) \cdot q} = \bigl(\alpha^{p^d - 1}\bigr)^q = 1^q = 1. \end{align*} Multiplying by $\alpha$: $\alpha^{p^n} = \alpha$, so $\alpha$ is a root of $t^{p^n} - t$. We have shown that every root of $t^{p^d} - t$ in $\overline{\mathbb{F}_p}$ is also a root of $t^{p^n} - t$. Since $t^{p^d} - t$ is separable (has $p^d$ distinct roots) and each of these roots appears among the roots of $t^{p^n} - t$, the polynomial $t^{p^d} - t$ divides $t^{p^n} - t$ in $\overline{\mathbb{F}_p}[t]$. Since both polynomials have coefficients in $\mathbb{F}_p$ and $t^{p^d} - t$ is monic (leading coefficient $1$), the quotient $(t^{p^n} - t)/(t^{p^d} - t)$ has coefficients in $\mathbb{F}_p$ as well (by polynomial long division), so $t^{p^d} - t \mid t^{p^n} - t$ in $\mathbb{F}_p[t]$. [/guided] [/step] [step:Conclude the forward direction: $d \mid n$ implies $\mathbb{F}_{p^d} \subset \mathbb{F}_{p^n}$] The elements of $\mathbb{F}_{p^d}$ are precisely the roots of $t^{p^d} - t$ in $\overline{\mathbb{F}_p}$, and the elements of $\mathbb{F}_{p^n}$ are precisely the roots of $t^{p^n} - t$ in $\overline{\mathbb{F}_p}$. Since every root of $t^{p^d} - t$ is a root of $t^{p^n} - t$ (by the previous step), every element of $\mathbb{F}_{p^d}$ is an element of $\mathbb{F}_{p^n}$, i.e., $\mathbb{F}_{p^d} \subset \mathbb{F}_{p^n}$. Moreover, $\mathbb{F}_{p^d}$ is a subfield of $\mathbb{F}_{p^n}$: it is closed under addition and multiplication (being a field in its own right) and contains $0$ and $1$ (as roots of $t^{p^d} - t$), so it inherits the field operations of $\mathbb{F}_{p^n}$ by restriction. This completes the forward implication. [/step] [step:Prove the converse: $\mathbb{F}_{p^d} \subset \mathbb{F}_{p^n}$ implies $d \mid n$] Assume $\mathbb{F}_{p^d}$ is a subfield of $\mathbb{F}_{p^n}$. Then $\mathbb{F}_{p^n}$ carries the structure of a vector space over $\mathbb{F}_{p^d}$, with scalar multiplication given by field multiplication in $\mathbb{F}_{p^n}$. Let $m := \dim_{\mathbb{F}_{p^d}} \mathbb{F}_{p^n}$, the dimension of $\mathbb{F}_{p^n}$ as a vector space over $\mathbb{F}_{p^d}$. Since $\mathbb{F}_{p^n}$ is a finite-dimensional vector space over the finite field $\mathbb{F}_{p^d}$ (which has $p^d$ elements), the cardinality of $\mathbb{F}_{p^n}$ equals the number of distinct $\mathbb{F}_{p^d}$-linear combinations of a basis. A basis has $m$ elements, each coefficient ranges over $\mathbb{F}_{p^d}$ (which has $p^d$ choices), so \begin{align*} |\mathbb{F}_{p^n}| = |\mathbb{F}_{p^d}|^m = (p^d)^m = p^{dm}. \end{align*} Since $|\mathbb{F}_{p^n}| = p^n$, we obtain $p^n = p^{dm}$. Comparing exponents (using the uniqueness of prime factorisation, or simply that the function $k \mapsto p^k$ is injective on $\mathbb{N}$) gives $n = dm$, so $d \mid n$. [guided] For the converse, we use a linear-algebraic argument. The hypothesis is that $\mathbb{F}_{p^d}$ is a subfield of $\mathbb{F}_{p^n}$, meaning $\mathbb{F}_{p^d} \subset \mathbb{F}_{p^n}$ and the field operations agree. Any field extension $F \subset E$ makes $E$ a vector space over $F$ (with vector addition being field addition in $E$, and scalar multiplication being field multiplication restricted to $F \times E$). In our case, $\mathbb{F}_{p^n}$ is a vector space over $\mathbb{F}_{p^d}$. Why is this vector space finite-dimensional? Because $\mathbb{F}_{p^n}$ is a finite set (it has $p^n$ elements), and any vector space over a field with $p^d \ge 2$ elements that has finitely many vectors must be finite-dimensional. (If the dimension were infinite, the space would contain infinitely many distinct vectors.) Let $m := \dim_{\mathbb{F}_{p^d}} \mathbb{F}_{p^n}$. Choose a basis $\{e_1, e_2, \ldots, e_m\}$ of $\mathbb{F}_{p^n}$ over $\mathbb{F}_{p^d}$. Every element of $\mathbb{F}_{p^n}$ can be written uniquely as \begin{align*} a_1 e_1 + a_2 e_2 + \cdots + a_m e_m, \quad a_i \in \mathbb{F}_{p^d}. \end{align*} Each coefficient $a_i$ can take $|\mathbb{F}_{p^d}| = p^d$ values independently, so the total number of elements is \begin{align*} |\mathbb{F}_{p^n}| = (p^d)^m = p^{dm}. \end{align*} Since $|\mathbb{F}_{p^n}| = p^n$, we conclude $p^n = p^{dm}$, hence $n = dm$ (since $p \ge 2$ and the exponential function $k \mapsto p^k$ is strictly increasing on the positive integers). Therefore $d \mid n$. Note that this argument also determines the degree of the extension: $[\mathbb{F}_{p^n} : \mathbb{F}_{p^d}] = m = n/d$. [/guided] [/step]