This property is mentioned in http://en.wikipedia.org/wiki/Aleph_number
I cannot find a contradiction assuming otherwise. Maybe this is proved by transfinite induction?
$\aleph_\alpha$ is defined to be the least infinite cardinal not in $\{\aleph_\beta\}_{\beta\in\alpha}$.
Edit The useful hint that allowed me to see the proof is the order-preserving injection from $\alpha$ to $\aleph_\alpha$. To explain, for future viewers, recall the following fact
Let $A$ and $B$ be sets such that $A\subseteq B$. Let $\prec_B$ be a well-order in $B$. If $\alpha$ is the ordinal of $(A,\prec_B^\ast)$ and $\beta$ is the ordinal of $(B,\prec_B)$, then $\alpha\le \beta$.
The order-preserving injection $f:\beta\mapsto \aleph_\beta$ (domain $\alpha$) makes the range (w/ ordinal $\alpha$) a subset of $\aleph_\alpha$ (w/ ordinal itself). Therefore $\alpha\le\aleph_\alpha$.
You should be able to find an injective function from $\alpha$ to $\aleph_{\alpha}$