An urn contains N1 white and N2 black balls from which k are drawn without replacement. Find the probability that the next ball drawn is white. (Please see the image attached) 
I tried solving the problem but the final expression I fail to simplify. I basically considered each possible cases i.e., drawing 0 white and k black, 1 white and (k-1) black etc and then multiplied by the probability of drawing a white on the next trial. I don't see how we can solve it otherwise. The book provided a solution that doesn't make sense to me. How can it be that simple?
 A: The counting of "event points" in the solution uses the fact that we can use something we know about what happened at the time a particular ball was drawn to deduce what could have happened at the time another particular ball was drawn.
For the "total number" we know that there are $N_1+N_2$ balls in the urn when the first one is drawn.
Once we know which of those balls was drawn first, there remain $N_1+N_2-1$ possibilities for which ball was drawn second.
(Each "first ball" gives us a different set of $N_1+N_2-1$ possibilities for the second ball, but we are just counting the total number of possibilities, so what matters is that there are always $N_1+N_2-1$ of them.)
For each of the $(N_1+N_2)(N_1+N_2-1)$ total possibilities for the first two balls, there are $N_1+N_2-2$ possibilities for the third, and so forth.
And that's how we end up with
$$ T = (N_1+N_2)(N_1+N_2-1)(N_1+N_2-2)\cdots(N_1+N_2-k)$$
event points in the entire probability space.
But this kind of counting argument actually depends only on the order in which we ask which ball was drawn at each time, not in the order in which the balls actually were drawn. We get the same total $T$ if we first consider the $k+1$st ball to be drawn, and then consider the first ball drawn, then the second, and so forth.
To count the events in which the $k+1$st ball is white, the book's solution considers the balls in just such a sequence.
There are $N_1$ ways in which the $k+1$st ball could be white.
For each of these, there are now just $N_1+N_2-1$ possibilities for the first ball drawn (since we already know that a particular ball was not drawn before the $k+1$st drawing), $N_1+N_2-2$ possibilities for the second ball, and so forth.
The count of possible event points is still a product of the number of choices available for each ball,
and the author has chosen to write these terms in the order in which the balls were drawn rather than the order in which the counting method identified each ball.
So the $k+1$st ball, although we "observed" its identity first, appears as the last term $N_1$ of the product:
$$ S = (N_1+N_2-1)(N_1+N_2-2)\cdots(N_1+N_2-k)N_1.$$
By symmetry, we assign identical probability to each of the original $T$ event points, and therefore the probability we are looking for is just $S/T,$ that is,
$$\frac ST = \frac{(N_1+N_2-1)(N_1+N_2-2)\cdots(N_1+N_2-k)N_1}
                  {(N_1+N_2)(N_1+N_2-1)(N_1+N_2-2)\cdots(N_1+N_2-k)}.$$
Look at the ratio on the right. Each term from $(N_1+N_2-1)$ to $(N_1+N_2-k)$ occurs exactly once on top and once on the bottom,
so all these terms can be canceled out. The only terms that cannot be canceled are $N_1$ on top and $N_1+N_2$ on the bottom.
So we end up with
$$\frac ST = \frac{N_1}{N_1+N_2}.$$

As noted in other comments and answer(s), this is not the easy way to find the probability. The easy way is just to imagine what happens if you keep your eyes shut while you draw the first $k$ balls and only open them when you are holding the $k+1$st ball, so that you see it before all the others.
You then have a somewhat complicated way of drawing a random ball from an urn, consisting of ejecting $k$ balls without looking before you make your final selection.
Every ball in the original urn is equally likely to be the ball selected in this way.
A: Draw the first ball and don't look at it.  The probability it is white is $\frac {N1}{N1+N2}$.  Now draw $k$ more balls and put them in a line after the first one.  Finally, swap the first and last balls.  The probability it is white has not changed.  This is quite general.  The fact that you drew $k$ balls does not matter as long as you don't look at them.
