By the Binomial Theorem, when $y<1$:
$(1+x)^y=1+yx+\dfrac{y(y-1)}{2!}x^2+\dfrac{y(y-1)(y-2)}{3!}...$
The coefficient of $y$ is:
$x+\dfrac{(-1)}{2!}x^2+\dfrac{(-1)(-2)}{3!}x^3+\dfrac{(-1)(-2)(-3)}{4!}x^4...$
I don't understand how the coefficient of $y$ came about except for the second term, which is obviously $x$ by inspection. For example in the third term, how is it possible to say the coefficient of $y$ in $y(y-1)$ is $-1$?
This is because in $y(y-1)$ as you are looking for coefficient of $y$ you have to keep in mind that what will get multiplied to give $y$. For eg in $y(y-1)$ expanding it you will realise that when you multiply $y $ with $y$ you get $y^2$ WHICH CLEARLY WON'T CONTRIBUTE TO COEFFICIENT OF $y$.. BUT when we multiply y with $-1$ the power of $y$ will remain as it is and so it will contribute to its coefficients. Now similarly in the next term $y(y-1)(y-2)$ you will have only $1$ term of y which will be formed only when we multiply $y.(-1).(-2)$ . We don't care about other terms as we know they will give higher powers of $y$ which won't contribute to its coefficients.