Showing $\kappa^{cf(\kappa)}$ determines the map $\kappa\mapsto 2^\kappa$

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Let $\kappa$ be a limit cardinal. I want to show

1) If $2^\alpha$ is eventually constant as $\alpha\rightarrow\kappa$, then $2^\kappa=2^{<\kappa}\cdot\kappa^{cf(\kappa)}$.

2) If $2^\alpha$ is not constant as $\alpha\rightarrow\kappa$, $2^\kappa=\lambda^{cf(\lambda)}$ where $\lambda=2^{<\kappa}$

Here $2^{<\kappa}$ denotes $\sum_{\alpha<\kappa}2^\alpha$. (Also, is $\sum_{\alpha<\kappa}2^\alpha=\sum_{\alpha<cf(\kappa)}2^\alpha$ here or not?) For part 1) I think that if $2^\alpha$ is eventually constant as $\alpha\rightarrow\kappa$, then $2^\kappa$ is that constant value, but I don't know where to go from there. If $\kappa$ is singular then $\kappa=\sum_{\nu<cf(\kappa)}\kappa_\nu$ with $\kappa_\nu<\kappa$, but I'm stuck. Thanks in advance.