Let $K$ be a number field with $r_1$ real and $r_2$ pairs of complex embeddings. Set $r = r_1 + r_2 - 1$.

Step 1: Logarithmic embedding and hyperplane

Define $\lambda: K^* \to \mathbb{R}^{r_1+r_2}$ by: $\lambda(\alpha) = (\log|\sigma_1(\alpha)|, \ldots, \log|\sigma_{r_1}(\alpha)|, 2\log|\tau_1(\alpha)|, \ldots, 2\log|\tau_{r_2}(\alpha)|)$

For units $\epsilon \in \mathcal{O}_K^*$, we have $|N(\epsilon)| = 1$, so: $\sum_{i=1}^{r_1}\log|\sigma_i(\epsilon)| + \sum_{j=1}^{r_2}2\log|\tau_j(\epsilon)| = 0$

Thus $\lambda(\mathcal{O}_K^*)$ lies in the hyperplane $H = \{x : \sum x_i = 0\} \cong \mathbb{R}^r$.

Step 2: Compactness argument

Consider the image of roots of unity: $\lambda(\mu_K) = \{0\}$ since $|\zeta| = 1$ for all roots of unity $\zeta$.

For $\epsilon \in \mathcal{O}_K^*$ not a root of unity, infinitely many powers $\epsilon^n$ are distinct. Their images $\lambda(\epsilon^n) = n\lambda(\epsilon)$ accumulate, showing $\lambda(\epsilon) \neq 0$.

Step 3: Discreteness of unit lattice

The set $\lambda(\mathcal{O}_K^*/\mu_K)$ is discrete in $H$. To see this, suppose $\{\epsilon_n\}$ are units with $\lambda(\epsilon_n) \to 0$.

Choose representatives with $|\sigma_i(\epsilon_n)| \geq 1$ for some fixed $i$. Then all other $|\sigma_j(\epsilon_n)|$ are bounded (by the trace-zero condition). Since $\mathcal{O}_K$ is discrete in the Minkowski embedding, only finitely many such $\epsilon_n$ exist modulo $\mu_K$.

Step 4: Fundamental domain

Consider the region: $F = \{x \in H : 0 \leq x_i < C \text{ for } i = 1, \ldots, r\}$

for sufficiently large $C$. This is a fundamental domain for the action of $\lambda(\mathcal{O}_K^*/\mu_K)$ on $H$.

The quotient $H / \lambda(\mathcal{O}_K^*/\mu_K)$ is compact, showing $\lambda(\mathcal{O}_K^*/\mu_K)$ is a full-rank lattice in $H \cong \mathbb{R}^r$.

Step 5: Freeness and rank

A discrete subgroup of $\mathbb{R}^r$ with compact quotient is a free abelian group of rank $r$. Therefore: $\mathcal{O}_K^*/\mu_K \cong \mathbb{Z}^r$

Taking generators $\epsilon_1, \ldots, \epsilon_r$ gives the fundamental units.

Step 6: Uniqueness of representation

If $\zeta \epsilon_1^{a_1} \cdots \epsilon_r^{a_r} = \zeta' \epsilon_1^{b_1} \cdots \epsilon_r^{b_r}$ with $\zeta, \zeta' \in \mu_K$, then: $\epsilon_1^{a_1-b_1} \cdots \epsilon_r^{a_r-b_r} = \zeta' / \zeta \in \mu_K$

Since $\lambda$ is injective on $\mathcal{O}_K^*/\mu_K$ and $\lambda(\mu_K) = \{0\}$, we have $a_i = b_i$ for all $i$, hence $\zeta = \zeta'$.