 Cofinality of a power-set cardinal

For an infinite cardinal number $\kappa$, use cardinal arithmetic with the axiom of choice. If $\lambda =2^{\kappa }$ had cofinality $\mu \le \kappa$, choose cofinal ordinals ${\alpha }_i<\lambda$, $i<\mu$. Their cardinalities $a_i$ satisfy ${\sum }_i{a}_i=\lambda$. The König theorem for cardinal numbers gives $\lambda <{\prod }_i{a}_i^{+}\le {\lambda }^{\mu }\le (2^{\kappa }{)}^{\kappa }=\lambda$, a contradiction.