# Matrix exponential is continuous

I want to prove that the function $$\exp\colon M_n(\mathbb{C})\to \mathrm{GL}_n(\mathbb{C})$$ is continuous under standard matrix norm $$\lVert A\rVert=\sup_{\lVert x\rVert=1}\lVert Ax\rVert.$$ Wikipedia says that it follows from the inequality $$\lVert e^{X+Y}-e^X\rVert\leqslant \lVert Y\rVert e^{\lVert X\rVert}e^{\lVert Y\rVert},$$ and I understand why, but I don't quite follow how to get this inequality. Could somebody explain that?

Let $$p:M_n(\Bbb C )^k\to M_n(\Bbb C ),\quad (X_1,X_2,\ldots ,X_n)\mapsto X_1\cdot X_2\cdots X_n\tag1$$ the ordered product of a vector of matrices, and $$c_X:\{X,Y\}^k\to \{0,\ldots ,k\}\tag2$$ is a function that count the number of coordinates of a vector in $$\{X,Y\}^k$$ that are equal to $$X$$. Then we have that $$(X+Y)^k=\sum_{v\in \{X,Y\}^k}p(v)=\sum_{j=0}^k\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)\tag3$$ And if $$X$$ and $$Y$$ commute then $$\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)=\binom{k}{j}X^jY^{k-j}\tag4$$ Then from $$\mathrm{(3)}$$ we have that \begin{align*} \|(X+Y)^k-X^k\|&\leqslant \left\|\sum_{j=0}^k\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)-X^k\right\|\\ &=\left\|\sum_{j=0}^{k-1}\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}p(v)\right\|\\ &\leqslant \sum_{j=0}^{k-1}\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}\|p(v)\|\\ &\leqslant \sum_{j=0}^{k-1}\sum_{\substack{v\in\{X,Y\}^k\\c_X(v)=j}}\|X\|^j\|Y\|^{k-j}\\ &=\sum_{j=0}^{k-1}\binom{k}{j}\|X\|^j\|Y\|^{k-j}\\ &=\sum_{j=0}^{k-1}\binom{k}{j}\|X\|^j\|Y\|^{k-j}+\|X\|^k-\|X\|^k\\ &=(\|X\|+\|Y\|)^k-\|X\|^k\tag5 \end{align*} where in the third inequality we used implicitly the inequality $$\|AB\|\leqslant \|A\|\|B\|$$ for any square matrices $$A$$ and $$B$$. Then you have that \begin{align*} \|e^{X+Y}-e^X\|&=\left\|\sum_{k\geqslant 0}\frac{(X+Y)^k-X^k}{k!}\right\|\\ &\leqslant \sum_{k\geqslant 0}\frac{\|(X+Y)^k-X^k\|}{k!}\\ &\leqslant \sum_{k\geqslant 0}\frac{(\|X\|+\|Y\|)^k-\|X\|^k}{k!}\\ &=e^{\|X\|+\|Y\|}-e^{\|X\|}\\ &=e^{\|X\|}(e^{\|Y\|}-1)\tag6 \end{align*} And $$e^c-1\leqslant ce^c\iff \sum_{k\geqslant 0}\frac{c^{k+1}}{(k+1)!}\leqslant \sum_{k\geqslant 0}\frac{c^{k+1}}{k!}\tag7$$ clearly holds for $$c\geqslant 0$$. Then $$\mathrm{(6)}$$ and $$\mathrm{(7)}$$ prove your inequality.