Proving directly that $\aleph_{\alpha+1} \cdot \aleph_{\alpha+1} = \aleph_{\alpha+1}$ by recursion

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I have defined the alephs by induction : $\aleph_{\alpha+1}$ is by definition the smallest cardinal superior to $\aleph_{\alpha}.$ Is there a quick way to show that $\aleph_{\alpha+1} \cdot \aleph_{\alpha+1} = \aleph_{\alpha+1}$ assuming that $\aleph_{\alpha} \cdot \aleph_{\alpha} = \aleph_{\alpha}$ ?