= Solution
Put $f=[k_L:k]$. For an <unramified extension> the <ramification index> is one and $[L:K]=f$. The residue $\bar\alpha$ has degree $f$ over $k$. Thus $1,\alpha,\ldots,\alpha^{f-1}$ are linearly independent over $K$: a nonzero <linear dependence> can be divided by a <coefficient> of least <valuation> and reduced to a nonzero relation among the reduced powers. Hence $[K(\alpha):K]\ge f=[L:K]$, and $\alpha$ generates $L$.
The same argument gives the integral statement, which is stronger than merely generating the <field>. Write
$$
x=\sum_{j=0}^{f-1}c_j\alpha^j,\qquad c_j\in K.
$$
If $m=\min_j v_K(c_j)$, divide by $\pi_K^m$. Since $\pi_K$ remains a <uniformizer> in the <unramified extension> and the reduced powers form a <basis> of $k_L/k$, the resulting sum has nonzero residue. Therefore $v_L(x)=m$. In particular $x$ is integral if and only if every <coefficient> is integral. This proves <unramified residue generator integral basis>:
$$
\boxed{\mathcal O_L=\bigoplus_{j=0}^{f-1}\mathcal O_K\alpha^j
=\mathcal O_K[\alpha].}
$$
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