Prove that $\lVert \exp(A) - \exp(B) \rVert \leq \lVert A-B \rVert e^{\max\{\lVert A \rVert, \lVert B \rVert \}}$ , where A and B are square Matrices, exp(A) is the matrix exponential
I can see where the $\lVert A-B \rVert$ comes from, but can't figure out how to get the $ e^{max\{\lVert A \rVert, \lVert B \rVert\}} $. The definition for the matrix exponential we're using is the following:
$\exp(A)= \sum \frac{A^k}{k!}$.
 A: Partial answer
Define
$$F(t) = \exp((1-t)A + tB)$$ for $t\in [0,1]$. According to the Mean Value Theorem
$$\Vert \exp(A) - \exp(B) \Vert = \Vert F(1) - F(0) \Vert \le \sup_{t \in [0,1]} \Vert F^\prime(t) \Vert \vert 1 - 0 \vert =\sup_{t \in [0,1]} \Vert F^\prime(t) \Vert$$
As $$F^\prime(t) = \exp((1-t)A + tB) \cdot (B-A)$$ we're left to prove that
$$\sup_{t \in [0,1]} \Vert \exp((1-t)A + tB) \Vert \le e^{\max(\Vert A \Vert,\Vert B \Vert)}$$ which is clear if $A,B$ commutes as in this case
$$\exp((1-t)A + tB) = \exp((1-t)A) \exp(tB).$$
If not, I don't see how to conclude...
A: You can start by restating the following inequality.
Lemma:
$\forall n \in \mathbb{N}$, we have:
$$ \| A^n -B^n \| \le n \|A-B\| c^{n-1}$$
where $c= \max( \| A \|, \| B\| )$.
Demonstration (Outline) 
It's just a direct consequence of the submultiplicity of $\| \|$. For example, for $n=3$ , $$ A^3-B^3 = (A-B)A^2+B(A-B)A+B^2(A-B)$$
$\square$
Then you can see that:
$$ \| \exp(A) -\exp(B) \| \le \sum_{n \ge 1} \| A^n-B^n \| \dfrac{1}{n!}$$
The rest is simple $\square$ 
Remark: A similar proof by MVT like given by mathecounterexample can also be carried out, some final arguments must ofcourse be completed by using the submultiplicity.
