Norm of difference in exponential of matrices

You can first get $\|B^k-A^k\|\leq k\|B-A\|(\max(\|A\|,\|B\|))^{k-1}$ as follows

\begin{align*} \|B^k-A^k\| &= \left\lVert\sum_{l=0}^{k-1}B^{l}(B-A)A^{k-1-l}\right\rVert \leq \sum_{l=0}^{k-1}\left\lVert B^{l}(B-A)A^{k-1-l}\right\rVert\\ &\leq \sum_{l=0}^{k-1}\|B\|^l\|B-A\|\|A\|^{k-1-l} \leq k\|B-A\|\left(\max(\|A\|,\|B\|)\right)^{k-1} \end{align*}

Then\begin{align*}\|e^B-e^A\|=\left\lVert \sum_{k=0}^\infty \frac{B^k-A^k}{k!}\right\rVert\leq \sum_{k=0}^\infty \frac{k\left(\max(\|A\|,\|B\|)\right)^{k-1}\|B-A\|}{k!}=\left|B-A\right|e^{\max(\|A\|,\|B\|)}\end{align*}