These [tex]N[/tex] outcomes make up the entire sample space, so
[tex]\displaystyle \sum_{k=1}^N P(e_k) = P(e_1) + P(e_2) + P(e_3) + \cdots + P(e_N) = 1[/tex]
We're given that [tex]P(e_{j+1}) = 2 P(e_j)[/tex] for all [tex]j\in\{1,2,\ldots,N-1\}[/tex], so
[tex]P(e_1) + 2 P(e_1) + 2^2 P(e_1) + \cdots + 2^{N-1} P(e_1) = 1 \\\\ \implies P(e_1) = \dfrac1{1 + 2 + 2^2 + \cdots + 2^{N-1}} = \dfrac1{2^N - 1}[/tex]
Then we can solve the recurrence relation to get the probability of the [tex]j[/tex]-th outcome,
[tex]P(e_{j+1}) = 2 P(e_j) = 2^2 P(e_{j-1}) = 2^3 P(e_{j-2}) = \cdots \\\\ \implies P(e_{j+1}) = 2^j P(e_1) \\\\ \implies P(e_j) = 2^{j-1} P(e_1) = \dfrac{2^{j-1}}{2^N - 1}[/tex]
The probability of getting this sequence of [tex]k[/tex] outcomes is then
[tex]\displaystyle P(E_k) = P(e_1) + P(e_2) + \cdots + P(e_k) = \sum_{j=1}^k \frac{2^{j-1}}{2^N-1} = \frac{2^k-1}{2^N-1}[/tex]
as required.
Some preliminary results: If [tex]S[/tex] is the sum of the first [tex]n[/tex] terms of a geometric series with first term [tex]a[/tex] and common ratio [tex]r[/tex], then
[tex]S = a + ar + ar^2 + \cdots + ar^{n-1}[/tex]
[tex]\implies rS = ar + ar^2 + ar^3 + \cdots + ar^n[/tex]
[tex]\implies S - rS = a(1 - r^n)[/tex]
[tex]\implies S = \dfrac{a(1 - r^n)}{1 - r}[/tex]
which gives us, for instance,
[tex]1 + 2 + 2^2 + \cdots + 2^{N-1} = \dfrac{1 - 2^N}{1 - 2} = 2^N-1[/tex]