Weak cardinal powers and singular cardinals Suppose $\kappa > \operatorname{cf}(\kappa)$. Show that:
i) if $\kappa$ strong limit then $\kappa^{<\kappa} = \kappa^{\operatorname{cf}(\kappa)}$
ii) if $\kappa$ not strong limit then $2^{<\kappa} = \kappa^{<\kappa} > \kappa$.
My thoughts:
for i), write $\kappa = \sum_{i < \operatorname{cf}(\kappa)} \kappa_i$ with $\kappa_i < \kappa$ for each i. Then, $\kappa^{<\kappa} = \sum_{\lambda < \kappa} \kappa^{\lambda} = \sum_{i<\operatorname{cf}(\kappa)}\kappa^{\kappa_i}$. Now, we know that $\kappa^{\lambda} \leq (\sum_{\alpha < \kappa} \alpha^{\lambda})^{\operatorname{cf}(\kappa)}$, so $\sum_{i<\operatorname{cf}(\kappa)}\kappa^{\kappa_i} \leq \sum_{i<\operatorname{cf}(\kappa)}(\sum_{\alpha < \kappa} \alpha^{\kappa_i})^{\operatorname{cf}(\kappa)}$, but I'm not sure what to do now.
for ii) I guess an argument similar to the above shows that $\kappa^{<\kappa}= \kappa^{\operatorname{cf}(\kappa)} > \kappa$, but I'm stuck on showing $2^{<\kappa} = \kappa^{<\kappa}$. How does merely knowing the fact that for some $\mu < \kappa$ $2^{\mu} \geq \kappa$ let us deduce the equivalence?
 A: For (i), notice that you have $\alpha^{\kappa_i}<\kappa$ in the notation of your last inequality. This is because $\kappa$ is a strong limit, so $\alpha^{\kappa_i}\leq (2^\alpha)^{\kappa_i}<\kappa$. But then, following your inequality,
$$\kappa^\lambda\leq \sum_{i<\mathrm{cf}(\kappa)}\left(\sum_{\alpha<\kappa}\alpha^{\kappa_i}\right)^{\mathrm{cf}(\kappa)}
\leq \sum_{i<\mathrm{cf}(\kappa)}\kappa^{\mathrm{cf}(\kappa)} = \kappa^{\mathrm{cf}(\kappa)}$$
Sending $\lambda$ to $\kappa$ we get $\kappa^{<\kappa}\leq \kappa^{\mathrm{cf}(\kappa)}$ and the reverse inequality is obvious.
For (ii), Brian and I have already argued why $2^{<\kappa}=\kappa^{<\kappa}$. Of course, since $\kappa$ is assumed to not be a strong limit, $2^{<\kappa}>\kappa$ must clearly hold. 
A: (ii) If $\kappa$ is not a strong limit, then $\kappa\le 2^\lambda$ for some $\lambda<\kappa$. Thus, if $\lambda\le\mu<\kappa$, then
$$2^\mu=\left(2^\lambda\right)^\mu\ge\kappa^\mu\ge 2^\mu\;,$$
and it follows that $2^{<\kappa}=\kappa^{<\kappa}$.
