So I tried proving it, and I think its kind of a natural proprety...however when looking for transposes I didn't find it.
So, for $(e^{At})^T=e^{A^Tt}$ to be equal, this is equivalent as saying $(\sum_{k=0}^\infty \frac {(At)^k}{k!})^T= \sum_{k=0}^\infty \frac {(A^Tt)^k}{k!}$. However, by transpose properties $(\sum_{k=0}^\infty \frac {(At)^k}{k!})^T=\sum_{k=0}^\infty \frac {t^k}{k!}(A^k)^T.$ Hence, one only needs to prove that $(A^k)^T=(A^T)^k$.
For this, lets use induction. Where we need two initial cases: $k=1$ which is trivial since $(A^1)^T=A^T=(A^T)^1$. And $k=2$ where $(AA)^T=A^TA^T=(A^T)^2$.
So les suppose $(A^k)^T=(A^T)^k$ for every $k\leq n$. Lets see if $n$ is odd. Then there exist $k\leq n$ such $n+1=2k$ and so $$(A^{n+1})^T=(A^k A^k)^T=(A^k)^T(A^k)^T=(A^T)^k(A^T)^k=(A^T)^{n+1}.$$
On the other hand, if $n$ is even, then there exist $k\leq n$ such that $n+1=2k+1$ then $$(A^{n+1})^T=(A^{2k}A)^T=A^T(A^kA^k)^T=A^T(A^k)^T(A^k)^T=A^T(A^T)^k(A^T)^k=(A^T)^{n+1}. $$
Are there any mistakes? I am scared I proved something false.
Your argument's right, but unnecessarily complicated, as we don't need to split the inductive step by parity. To go from $n=k$ to $n=k+1$, use$$(A^{k+1})^T=(A^kA)^T=A^T(A^k)^T=A^T(A^T)^k=(A^T)^{k+1}.$$While the first and second expressions of the form $X^{k+1}$ are respectively written as $X^kX$ and $XX^k$, both are correct because square matrices are power-associative.