Exercise 18, § 6

(a) For each ordinal $\alpha$ and each cardinal $\mathfrak{m} \neq 0$ we have $\aleph_{\alpha+1}^{\mathfrak{m}} = \aleph_\alpha^{\mathfrak{m}} \cdot \aleph_{\alpha+1}$ (reduce to the case where $\mathfrak{m} < \aleph_{\alpha+1}$ and consider the mappings of the cardinal $\mathfrak{m}$ into the ordinal $\omega_{\alpha+1}$).

(b) Deduce from (a) that, for each ordinal $\gamma$ such that Card $(\gamma) \leqslant \mathfrak{m}$, we have $\aleph_{\alpha+\gamma}^{\mathfrak{m}} = \aleph_\alpha^{\mathfrak{m}} \cdot \aleph_{\alpha+\gamma}^{\mathrm{Card}(\gamma)}$ (by transfinite induction on $\gamma$).

(c) Deduce from (b) that, for each ordinal $\alpha$ such that Card $(\alpha) \leqslant \mathfrak{m}$, we have $\aleph_\alpha^{\mathfrak{m}} = 2^{\mathfrak{m}} \cdot \aleph_\alpha^{\mathrm{Card}\,(\alpha)}$.