If $X_1,X_2,\ldots,X_n$ are independent and identically distributed continuous random variables, show that $P(X_1<X_2,X_1<X_3,\ldots,X_1<X_n)=\frac{1}{n}$.
So far, this is what I have: $$P(X_1<X_2,X_1<X_3,\ldots,X_1<X_n)=\int_{-\infty}^{\infty}P(X_1<X_2,X_1<X_3,\ldots,X_1<X_n|X_1=x)\cdot f_{X_1}(x)dx$$ $$=\int_{-\infty}^{\infty}P(X_1<x,X_1<x,\ldots,X_1<x)\cdot f_{X_1}(x)dx$$ $$=\int_{-\infty}^{\infty}\prod_{i=1}^{n}P(X_i>x)\cdot f_{X_1}(x)dx$$
This is where I'm stuck. I know that since they're all identically distributed, there is probably some way to put everything together that is inside the integral, but I'm having a hard time figuring out how $P(X_i>x)$ relates to the distribution of $X_1$.