Suppose $\alpha$ is a cardinal.
Suppose $\beta\geq cf(\alpha)$, and let $f: \beta \to \alpha$ be unbounded. Then $\alpha = \displaystyle\bigcup_{\delta\in \beta}f(\delta) = |\displaystyle\bigcup_{\delta\in \beta} f(\delta) | \leq \displaystyle\sum_{\delta \in \beta} |f(\delta)|$.
For $\delta \in \beta$, $f(\delta) <\alpha$ and since $\alpha$ is a cardinal each of the $f(\delta)$ is of cardinality strictly less than $\alpha$ so we have found a partition as desired.
Converseley, suppose $\alpha = \displaystyle\sum_{\delta \in \beta}\kappa_\delta$ is a partition and $\kappa_\delta$ is a cardinal strictly less than $\alpha$ and $\beta<\alpha$ ( we may assume $\beta <\alpha$ otherwise $\beta \geq \alpha$ and $\beta \geq cf(\alpha)$ follows immediately. We may also assume $\beta$ is a cardinal)
Consider then the family $\beta \to \alpha$, $\delta \mapsto \kappa_\delta$. If it were bounded below $\alpha$, say by $\mu$, then the cardinal of $\kappa_\delta$ would be $\leq |\mu|$, and the cardinal of the sum would be bounded by $\beta|\mu| = \max \{\beta, |\mu|\} <\alpha$, a contradiction. So this family is unbounded, and so $\beta \geq cf(\alpha)$, as desired.
We proved $\beta \geq cf(\alpha)$ if and only if there exists a partition of $\alpha$ in $\beta$ parts of cardinality $<\alpha$, so $cf(\alpha)$ is the least cardinal with this property.
