Distribution of the index of the variable which achieves the minimum of exponential random variables

it is said that the index of the variable which achieves the minimum is distributed according to the law $$P(k\mid X_k=\min\{X_1,X_2,...,X_n\})~=~\frac{\lambda_k}{\lambda_1+...+\lambda_n}$$

Y...eah.   That's a poor way to express it.   That is not actually a conditional probability.  

All they are saying is if we let $K$ be the random variable defined as the index of the minimum value of the sample, that is $X_K=\min\{X_1,X_2,\ldots,X_n\}$, then the probability mass function of $K$ is : $$P(K=k) ~=~ \dfrac{\lambda_k}{\lambda_1+\lambda_2+\cdots+\lambda_n}~~\mathbf 1_{k\in\{1,2,\ldots,n\}}$$

So, for $n=2$ you have found $P(X_1\leqslant X_2)~=~ P(K=1) ~=~ \dfrac{\lambda_1}{\lambda_1+\lambda_2}$

In general $$\begin{align}P(K=k) ~&=~ \int_0^\infty f_{X_k}(t) \prod\limits_{j\in\{1..n\}\setminus\{k\}} (1-F_{X_j}(t))\operatorname d t \\[1ex] &\vdots\\[1ex] &=~ \dfrac{\lambda_k}{\sum_{j=1}^n \lambda_j}\end{align}$$