Waiting time for a better offer This exercise is form "A Second Course in Stochastic Processes, Band 2" - Karlin.

Let $(X_n)_{n\in\mathbb{N}}$ be a family of i.i.d. random variables. Define $N:=\inf \{n\in\mathbb{N}: X_n>X_1\}$.
Determine $P[N=k]$ and $E[N]$.

Although this exercise is from a stochastic processes book, I have no experience in this field. However, I still wanted to try.
My attempt:
As the random variables are independent, I get
$$P[N=k]=P[X_2\le X_1]P[X_3\le X_1]...P[X_{k-1}\le X_1]P[X_k>X_1]\\ =\prod_{i=2}^{k-1}P[X_i\le X_1]\cdot P[X_k > X_1].$$
Now I use that the family is identically distributed and get
$P[N=k]=P[X_2\le X_1]^{k-2}\cdot P[X_k>X_1].$
Unfortunately, here I am stuck and don't know if the approach is correct.
 A: $$ P(N>k) = P(\bigcap_{i=2}^{k} X_i \leq X_1) = \mathbb E [P(\bigcap_{i=2}^k X_i\le X_1|X_1)] = \mathbb E [\prod_{i=2}^kP(X_i\le X_1|X_1)] = \mathbb E [F(X_1)^{k-1}] $$
where $F$ denotes the CDF of $X_1$.
Hence
$$ \mathbb E N = \sum_{k\geq 0} P(N>k) = 1+ \sum_{k=1}^\infty P(N>k) = 1 + \sum_{k=1}^\infty \mathbb EF[(X_1)^{k-1} ]  $$
Now since $F(X_1) \sim U[0,1]$ we have
$$ \mathbb E[F(X_1)^k] = \int_0^1 x^k dx = \frac{1}{k+1}  $$
and so \begin{align*} P(N=1)&=0 \\ P(N=k) &= P(N>k)-P(N>k-1) = \frac{1}{k}-\frac1{k-1}\\ \mathbb E N &= \infty \end{align*}
EDIT: the previous case is when $F$ is continuous! 
If it is not, then: if the support of $X_1$ is finite, there is positive probability that $X_1$ assumes the maximum value, thus on this event $N=\infty$, hence $EN =\infty$. It is a bit tricky to compute the probabilities, since $F$ has jumps and it does not have a proper inverse. Note that
$$P (F(X_1)<x ) =\sum_{k:F(k)< x} P(X_1=k).$$
The usual way to define a pseudoinverse is
$$ F^{-1} (q)= \sup\{ x: F(x)< q \} $$
or
$$ \inf\{x: q< F(x)\}. $$ 
