I'm looking into using $2^\kappa = \kappa^\kappa$ to prove this because it follows from $2 < \kappa < \kappa^+$ but I'm not sure how to connect the two together. Anyway, this might not even be the right approach to this problem.
Only a hint on how to show this is true - if not a complete proof - would also be great. I'm just asking out of curiosity, by the way. This isn't my homework.
In $\sf ZFC$ one can argue that since every two cardinals are comparable, either $\kappa^+\leq2^\kappa$, or $2^\kappa\leq\kappa^+$. However from Cantor's theorem $\kappa<2^\kappa$, and by definition we have $\kappa<\lambda\rightarrow\kappa^+\leq\lambda$, and this gives us the wanted proof. But we can prove some part of this without the axiom of choice, and then only use the axiom of choice for the final punch. Here are the details.
This is a combination of the following two theorems. One using the axiom of choice, and another not.
Proof. Note that $\kappa$ is equipotent with $\kappa\times\kappa$ (without an appeal to the axiom of choice, this is proved from the definition of Hessenberg sums for ordinals). We define the following map $\pi\colon\mathcal P(\kappa\times\kappa)\to\kappa^+$: $$\pi(A)=\begin{cases}\alpha & A\text{ is a well-ordering of its domain, and its order type is }\alpha\\0&\text{otherwise}\end{cases}$$
Since every ordinal $\alpha<\kappa^+$ can be mapped into $\kappa$, we have that every $\alpha<\kappa^+$ appears in the range of $\pi$, and therefore this is a surjection, as wanted. $\square$
Now to the part where the axiom of choice is invoked.
Proof. Suppose $f\colon A\to B$ is a surjection, then $\{f^{-1}(b)\mid b\in B\}$ is a family of non-empty sets. From the axiom of choice it follows that there is a function $g\colon B\to A$ such that $g(b)\in f^{-1}(b)$, as wanted. $\square$
Proof. There exists a surjection from $2^\kappa$ onto $\kappa^+$, and therefore an injection from $\kappa^+$ into $2^\kappa$. $\square$