Find density of the smallest value in a sample

Let $X_1, X_2,..., X_n$ be a random sample from an infinite population with density function $f(x)$ and distribution function $F(x).$ Let $Y_{(1)}$ be the smallest value in the sample (the first order statistic). Express the density of $Y_{(1)}$ in terms of $n, f,$ and $F.$ This is for a homework problem.

I know that I wanna use the distribution function method in order to solve this.

$P(Y_{(1)}\le y)= 1 -P(Y_{(1)}\gt y)$

$= 1 -P(X_1,X_2,...X_n\gt y)$

$= 1 -P(X_1\gt y)P(X_2\gt y)\ ...P(X_n\gt y)$

$= 1 - [F(y)]^n$

Now where do I go from here?

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You almost have it, except that the last line should read $1 - (1-F(y))^n$, because the probabilities $\Pr[X_i > y]$ are survival probabilities--note the direction of the inequality. Then take the derivative of the result, which gives you a density.
Note that we want $$1-\Pr(X_1\ge y)\Pr(X_2\ge y)\cdots \Pr(X_n\ge y).$$ Since our distribution is continuous, $\Pr(X_i=y)=0$, and therefore $$\Pr(X_i\ge y)=1-\Pr(X_i\le y)=1-F(y).$$ It follows that the cdf of our random variable is $1-(1-F(y))^n$.
Differentiate. By the Chain Rule, the density is $nf(y)(1-F(y))^{n-1}$.
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