Let $\lambda$ be a singular strong limit cardinal.
Prove that $2^\lambda = \lambda^{\mbox{cf}\lambda}$.
It has been a while since I had to prove anything relating to cardinals, and I am not sure where to start. A little push in the right direction would be very appreciated.
(Not homework)
Let $\langle \lambda_\xi\rangle_{\xi<\operatorname{cf}\lambda}$ be a cofinal sequence of $\lambda$. Since $\sum_{\xi<\operatorname{cf}\lambda}\lambda_\xi =\lambda$, we get $$ 2^\lambda= 2^{\sum\lambda_\xi}=\prod_{\xi<\operatorname{cf}\lambda}2^{\lambda_\xi} \le \prod_{\xi<\operatorname{cf}\lambda} \lambda = \lambda^{\operatorname{cf}\lambda} $$ since $2^{\lambda_\xi}<\lambda$ for all $\xi<\operatorname{cf}\lambda$. (Check that $\lambda$ is a strong limit.) In other direction, by König's lemma we get $$ \lambda^{\operatorname{cf}\lambda}=\left( \sum_{\xi<\operatorname{cf}\lambda} \lambda_\xi\right)^{\operatorname{cf}\lambda} \le \left( \prod_{\xi<\operatorname{cf}\lambda} 2^\lambda\right)^{\operatorname{cf}\lambda} =2^{\lambda\cdot\operatorname{cf}\lambda}. $$ It is easy to check that $\lambda_\xi<2^\lambda$ for all $\xi<\operatorname{cf}\lambda$ and $2^{\lambda\cdot\operatorname{cf}\lambda}=2^\lambda$.