[proofplan] Since $\mathbb{F}_{q^n}/\mathbb{F}_q$ is a Galois extension, the trace and norm are computed by summing and multiplying over the Galois group. By the [Galois Group of Finite Field Extension](/theorems/1277), $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q) = \langle \varphi \rangle$ where $\varphi$ is the Frobenius automorphism $x \mapsto x^q$, and the $n$ distinct $\mathbb{F}_q$-automorphisms are $\operatorname{id}, \varphi, \varphi^2, \ldots, \varphi^{n-1}$. The [Trace and Norm via Homomorphisms](/theorems/1294) theorem expresses the trace as a sum and the norm as a product over all $K$-homomorphisms. Substituting the explicit Frobenius iterates yields the stated formulas for the trace; for the norm, a geometric series computation converts the product of Frobenius iterates into a single power of $\alpha$. [/proofplan] [step:Identify the $\mathbb{F}_q$-automorphisms of $\mathbb{F}_{q^n}$ as iterates of the Frobenius] Define the Frobenius automorphism \begin{align*} \varphi \colon \mathbb{F}_{q^n} &\to \mathbb{F}_{q^n} \\ x &\mapsto x^q. \end{align*} By the [Galois Group of Finite Field Extension](/theorems/1277), the extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ is Galois of degree $n$, and \begin{align*} \operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q) = \{\operatorname{id}, \varphi, \varphi^2, \ldots, \varphi^{n-1}\}, \end{align*} a cyclic group of order $n$ generated by $\varphi$. For each $k \in \{0, 1, \ldots, n-1\}$, the automorphism $\varphi^k$ acts on $\alpha \in \mathbb{F}_{q^n}$ by \begin{align*} \varphi^k(\alpha) = \alpha^{q^k}. \end{align*} This follows by induction: $\varphi^0(\alpha) = \alpha = \alpha^{q^0}$, and if $\varphi^k(\alpha) = \alpha^{q^k}$, then $\varphi^{k+1}(\alpha) = \varphi(\alpha^{q^k}) = (\alpha^{q^k})^q = \alpha^{q^{k+1}}$, using the fact that $\varphi$ is a ring homomorphism (so it commutes with exponentiation). [guided] The Frobenius map $\varphi \colon x \mapsto x^q$ is an $\mathbb{F}_q$-automorphism of $\mathbb{F}_{q^n}$ because: - It is a field homomorphism: $\varphi(x + y) = (x + y)^q = x^q + y^q = \varphi(x) + \varphi(y)$ (the cross terms in the binomial expansion vanish since $q = p^k$ and $\binom{q}{j} \equiv 0 \pmod{p}$ for $0 < j < q$), and $\varphi(xy) = (xy)^q = x^q y^q = \varphi(x)\varphi(y)$. - It fixes $\mathbb{F}_q$ pointwise: for $a \in \mathbb{F}_q$, Fermat's Little Theorem in finite fields gives $a^q = a$ (since every element of $\mathbb{F}_q$ is a root of $t^q - t$). - It is injective: if $\varphi(x) = \varphi(y)$, then $x^q = y^q$, so $(x - y)^q = 0$, hence $x = y$ (as $\mathbb{F}_{q^n}$ is a field). Since $\mathbb{F}_{q^n}$ is finite, injectivity implies surjectivity, so $\varphi$ is an automorphism. The [Galois Group of Finite Field Extension](/theorems/1277) theorem guarantees that $\varphi$ has order exactly $n$ in $\operatorname{Gal}(\mathbb{F}_{q^n}/\mathbb{F}_q)$. The order cannot be less than $n$: if $\varphi^m = \operatorname{id}$ for some $m < n$, then every $\alpha \in \mathbb{F}_{q^n}$ would satisfy $\alpha^{q^m} = \alpha$, i.e., every element of $\mathbb{F}_{q^n}$ would be a root of the polynomial $t^{q^m} - t$. But this polynomial has at most $q^m$ roots, while $|\mathbb{F}_{q^n}| = q^n > q^m$, a contradiction. Since the Galois group has order $[\mathbb{F}_{q^n} : \mathbb{F}_q] = n$ and contains the element $\varphi$ of order $n$, the group is cyclic, generated by $\varphi$. The inductive computation of $\varphi^k(\alpha)$ is worth emphasising. We do not define $\varphi^k(\alpha) = \alpha^{q^k}$ as a notational convention — we derive it from the ring homomorphism property. Since $\varphi$ is multiplicative, $\varphi(\alpha^{q^k}) = (\alpha^{q^k})^q = \alpha^{q \cdot q^k} = \alpha^{q^{k+1}}$, and the induction closes. [/guided] [/step] [step:Apply the homomorphism formula for trace and norm] Since $\mathbb{F}_{q^n}/\mathbb{F}_q$ is a Galois extension of degree $n$, it is in particular a finite separable extension. The $n$ distinct $\mathbb{F}_q$-homomorphisms $\mathbb{F}_{q^n} \hookrightarrow \overline{\mathbb{F}_q}$ are precisely the elements of the Galois group: $\varphi^0 = \operatorname{id}, \varphi, \varphi^2, \ldots, \varphi^{n-1}$. (In a Galois extension, every $K$-homomorphism into an algebraic closure is in fact an automorphism of the extension field, since the extension is normal.) By the [Trace and Norm via Homomorphisms](/theorems/1294) theorem, for any $\alpha \in \mathbb{F}_{q^n}$: \begin{align*} \operatorname{Tr}_{\mathbb{F}_{q^n}/\mathbb{F}_q}(\alpha) &= \sum_{k=0}^{n-1} \varphi^k(\alpha) = \sum_{k=0}^{n-1} \alpha^{q^k} = \alpha + \alpha^q + \alpha^{q^2} + \cdots + \alpha^{q^{n-1}}. \end{align*} This establishes the trace formula. [guided] The [Trace and Norm via Homomorphisms](/theorems/1294) theorem requires the extension $L/K$ to be finite and separable, and expresses the trace and norm in terms of the $n = [L:K]$ distinct $K$-homomorphisms $\varphi_1, \ldots, \varphi_n \colon L \hookrightarrow \bar{K}$ as \begin{align*} \operatorname{Tr}_{L/K}(\alpha) &= \sum_{i=1}^{n} \varphi_i(\alpha), \\ \operatorname{N}_{L/K}(\alpha) &= \prod_{i=1}^{n} \varphi_i(\alpha). \end{align*} We verify the hypotheses. The extension $\mathbb{F}_{q^n}/\mathbb{F}_q$ has degree $n$, so it is finite. Every algebraic extension of a finite field is separable (by the [Finite Field Extensions Are Separable](/theorems/1268) theorem: inseparability would require $f(t) = g(t^p)$ for an irreducible $f$, but the surjectivity of Frobenius on $\mathbb{F}_q$ would then allow factoring $f$ as a $p$-th power, contradicting irreducibility). So both hypotheses are satisfied. The identification of $K$-homomorphisms with Galois group elements deserves a word. For a general separable extension $L/K$, the $K$-homomorphisms $L \hookrightarrow \bar{K}$ need not map $L$ to $L$; they are embeddings, not automorphisms. But for a Galois (= normal and separable) extension, normality ensures that every $K$-homomorphism $L \hookrightarrow \bar{K}$ has image contained in $L$, hence is an automorphism of $L$. Therefore the set of $K$-homomorphisms equals $\operatorname{Gal}(L/K) = \{\varphi^0, \varphi, \ldots, \varphi^{n-1}\}$. Substituting $\varphi^k(\alpha) = \alpha^{q^k}$ (established in the previous step) into the sum gives the trace formula directly. [/guided] [/step] [step:Compute the norm as a single power of $\alpha$ via the geometric series] Applying the product formula from the [Trace and Norm via Homomorphisms](/theorems/1294) theorem: \begin{align*} \operatorname{N}_{\mathbb{F}_{q^n}/\mathbb{F}_q}(\alpha) &= \prod_{k=0}^{n-1} \varphi^k(\alpha) = \prod_{k=0}^{n-1} \alpha^{q^k} = \alpha^{\sum_{k=0}^{n-1} q^k}. \end{align*} The last equality uses the law of exponents in the commutative ring $\mathbb{F}_{q^n}$: $\alpha^a \cdot \alpha^b = \alpha^{a+b}$, applied iteratively to the $n$ factors. The exponent is a finite geometric sum. Since $q \neq 1$ (as $q$ is a prime power, so $q \geq 2$), the geometric series formula gives \begin{align*} \sum_{k=0}^{n-1} q^k = \frac{q^n - 1}{q - 1}. \end{align*} Therefore \begin{align*} \operatorname{N}_{\mathbb{F}_{q^n}/\mathbb{F}_q}(\alpha) = \alpha^{1 + q + q^2 + \cdots + q^{n-1}} = \alpha^{(q^n - 1)/(q - 1)}, \end{align*} which completes the proof. [guided] The product-to-power conversion is the step that distinguishes the norm formula from the trace formula. While the trace is an irreducible sum (there is no closed-form simplification of $\alpha + \alpha^q + \cdots + \alpha^{q^{n-1}}$ in general), the norm collapses because multiplication of powers translates into addition of exponents: \begin{align*} \prod_{k=0}^{n-1} \alpha^{q^k} = \alpha^{q^0} \cdot \alpha^{q^1} \cdot \alpha^{q^2} \cdots \alpha^{q^{n-1}} = \alpha^{q^0 + q^1 + q^2 + \cdots + q^{n-1}}. \end{align*} This step is purely algebraic and holds in any commutative ring where $\alpha$ is a unit (or even when $\alpha = 0$, where both sides equal $0$, since the exponent $\sum_{k=0}^{n-1} q^k = 1 + q + \cdots + q^{n-1} \geq 1$). For the geometric series: we need $q \neq 1$ to apply the standard formula. Since $q = p^k$ for a prime $p$, we have $q \geq 2$, so the formula \begin{align*} 1 + q + q^2 + \cdots + q^{n-1} = \frac{q^n - 1}{q - 1} \end{align*} is valid. Note that the exponent $(q^n - 1)/(q - 1)$ is always a positive integer: the numerator $q^n - 1$ factors as $(q - 1)(q^{n-1} + q^{n-2} + \cdots + 1)$. As a consistency check: when $\alpha \neq 0$, we have $\alpha \in \mathbb{F}_{q^n}^*$, which is a cyclic group of order $q^n - 1$. The norm $\alpha^{(q^n-1)/(q-1)}$ has order dividing $(q^n - 1)/\bigl((q^n-1)/(q-1)\bigr) = q - 1$, so $\operatorname{N}(\alpha)^{q-1} = \alpha^{q^n - 1} = 1$. This confirms that $\operatorname{N}(\alpha) \in \mathbb{F}_q^*$, consistent with the general fact that the norm maps into the base field. When $\alpha = 0$, the norm equals $0^{(q^n-1)/(q-1)} = 0 \in \mathbb{F}_q$, which is also correct. [/guided] [/step]