$\exp(A)=\exp(B)\Rightarrow A,B$ similar

Question

Let $A,B\in\mathbb{C}^{n\times n}$. I want to show the following: If $\exp(A)=\exp(B)$ then $A$ and $B$ are similar. Here \begin{align} \exp:\mathbb{C}^{n\times n}\rightarrow \mathbb{C}^{n\times n}, A\mapsto\sum_{k=0}^\infty \frac{A^k}{k!} \end{align} denotes the matrix exponential map. I already proved the well-known properties: $\exp(TXT^{-1})=T\exp(X)T^{-1}$, if $T$ is invertible and $\exp(\operatorname{diag}(a_1,...,a_n))=\operatorname{diag}(\exp(a_1),...,\exp(a_n))$. Does someone have an idea from here?