$X_i$ are independent exponential distribution with parameter $\lambda_i$.
In Ross's textbook,
\begin{align} \mathbb{P}(X_{i_1}<\cdots<X_{i_n}\mid\min_i X_i>t)&=\mathbb{P}(X_{i_1}-t<\cdots<X_{i_n}-t\mid\min_i X_i>t) \\&=\mathbb{P}(X_{i_1}<\cdots<X_{i_n}) \end{align} I can't figure out how to get the second equality by the very definition of memoryless $$\mathbb{P}(X>s+t\mid X>t)=\mathbb{P}(X>s)$$ although it is true by intuition.
Edit: I have worked out this problem. $$\mathbb{P}(X>s+t\mid X>t)=\mathbb{P}(X>s)$$ implies $$\mathbb{P}(X>s+t)=\mathbb{P}(X>s)\mathbb{P}(X>t)$$ $d/ds$ on both sides: $$f_X(s+t)=f_X(s)\mathbb{P}(X>t)$$
Only prove $\mathbb{P}(X_1<X_2<X_3\mid X_1>t,X_2>t,X_3>t)$ for convenient.
\begin{align} &\mathbb{P}(X_1<X_2<X_3\mid X_1>t,X_2>t,X_3>t)\\ =&\frac{ \mathbb{P}( t<X_1<X_2<X_3)}{\mathbb{P}( X_1>t,X_2>t,X_3>t)}\\ =&\frac{ \int_{t<x_1<x_2}\mathbb{P}( t<x_1<x_2<X_3\mid X_1=x_1,X_2=x_2)f_{X_1}(x_1)f_{X_2}(x_2)dx_1dx_2}{\mathbb{P}( X_1>t,X_2>t,X_3>t)}\\ =&\frac{ \int_{t<x_1<x_2}\mathbb{P}(x_2<X_3)f_{X_1}(x_1)f_{X_2}(x_2)dx_1dx_2}{\mathbb{P}( X_1>t,X_2>t)\mathbb{P}(X_3>t)}\\ =&\frac{ \int_{t<x_1<x_2}\mathbb{P}(X_3>x_2-t)f_{X_1}(x_1)f_{X_2}(x_2)dx_1dx_2}{\mathbb{P}( X_1>t,X_2>t)}\\ =&\frac{ \int_{0<x_1<x_2}\mathbb{P}(X_3>x_2)f_{X_1}(x_1+t)f_{X_2}(x_2+t)dx_1dx_2}{\mathbb{P}( X_1>t)\mathbb{P}(X_2>t)}\\ =&\int_{0<x_1<x_2}\mathbb{P}(X_3>x_2)f_{X_1}(x_1)f_{X_2}(x_2)dx_1dx_2\\ =&\int_{0<x_1<x_2}\mathbb{P}(X_3>x_2>x_1|X_1=x_1,X_2=x_2)f_{X_1}(x_1)f_{X_2}(x_2)dx_1dx_2\\ =&\mathbb{P}(X_1<X_2<X_3) \end{align}
Hope someone check this proof.