I know that an infinite $\sigma$-algebra $\mathcal{A}$ has at least cardinality $\mathfrak{c}=2^{\aleph_0}$.
Suppose now that $\mathcal{A}$ is $\alpha$-generated ($\alpha>\omega$). Clearly, $\mathcal{A}$ is atomless. But what is its cardinality precisely?
Edit: By $\alpha$-generated I mean that the set of generators of $\mathcal{A}$ has size $\alpha$.
An $\alpha$-generated sigma-algebra $\mathcal A$ has $$ \alpha \le |\mathcal A| \le \alpha^{\aleph_0} $$ In case $\alpha=\alpha^{\aleph_0}$, this answers the question.
The fact you refer to in the question is: there is no sigma-algebra with $$ \aleph_0 \le |\mathcal A| < \mathfrak c = \aleph_0^{\aleph_0} $$ so we conclude that an infinite countably-generated sigma algebra has power exactly $\mathfrak c$.
Let $\mathcal A$ be a sigma-algebra on a set $X$ generated by $\mathcal G$ with $|\mathcal G| = \alpha$. Here is a proof for $|\mathcal A| \le \alpha^{\aleph_0}$
Let $\Omega$ be the set of all countable ordinals. We recursively define sets $\mathcal G_\lambda \subseteq \mathcal P(X)$ for $\lambda \in \Omega$ as follows.
$\bullet\;\mathcal G_0 = \mathcal G$
$\bullet\;\mathcal G_{\lambda+1}$ is the set of all countable unions $\bigcup_{n=1}^\infty E_n$ where, for each $n$, either $E_n \in \mathcal G_\lambda$ or $X\setminus E_n \in \mathcal G_\lambda$.
$\bullet\;$If $\lambda$ is a limit ordinal, then $$ \mathcal G_\lambda = \bigcup_{\gamma < \lambda}\mathcal G_\gamma . $$
Then: $$ \sigma(\mathcal G) = \bigcup_{\lambda \in \Omega}\mathcal G_\lambda $$ and we may prove recursively that $|\mathcal G_\lambda| \le \alpha^{\aleph_0}$ for all $\lambda \in \Omega$.
Next, $|\Omega| = \aleph_1 \le \mathfrak c = 2^{\aleph_0} \le \alpha^{\aleph_0}$, so $$ |\sigma(\mathcal G)| = \left|\bigcup_{\lambda \in \Omega}\mathcal G_\lambda\right| \le \aleph_1\cdot\alpha^{\aleph_0} \le \alpha^{\aleph_0}\cdot \alpha^{\aleph_0} = \alpha^{\aleph_0} . $$