Can't Derive Expectation in Second Price Auction I'm working on a SPA where we solve:

$\max_{\beta \ge 0} Pr\{Winning\}[v - \mathbb{E}(b^{[2]} \ | \ b^{[2]} \le \beta)]$

We assume all beliefs on bids are independently and identically distributed on $[0,1]$ given by CDF $F(b)$. 

$b^{[2]}$ denotes the second highest bid.

I've found $Pr\{Winning\} = Pr\{\beta \ge b_1, \ldots , b_n\} = F(b)^N$

In my lecture notes we are given:

$\mathbb{E}(b^{[2]} \ | \ b^{[2]} \le \beta) = \frac{1}{Pr(b^{[2]} \le beta)} \int_0^{\beta} bdF(b)^N$

I'm assuming this comes from $P(A|B) = \frac{P(A \cap B)}{P(B)} $, but can't quite figure it out. Especially the $F(b)^N$ in the integral.
There are $N+1$ bidders. So we only care about the second highest bid. That's the same as finding the highest bid amongst the remaining $N$ bidders. The expected second highest bid is given by: 

$$b \cdot Pr[b] \cdot Pr[b \geq b_{-i}] = b \cdot f(b) \cdot F(b)^{N-1} = b \cdot dF(b)^{N}$$

As expected value is an average, we divide out by the probability that $b \leq \beta$, which is $F(\beta)$.

Thus, we have:

$$\mathbb{E}[b^{[2]} | b^{[2]} \leq \beta] = \dfrac{1}{F(\beta)}\int_{0}^{\beta} bdF(b)^{N}$$