Strange sum in matrix theory I was doing some Matrix theory when a strange sum appered.
Exercise:
Suppose that $N$ is a nilpotent matrix of index $n$. Show that the equation 
$$X^2 = I + N$$ has a solution 
$$ X = \sum_{k=0}^{n-1} {\frac{1}{2}\choose{k}} N^k.$$
There was a hint to this exercise:
Because $(1+x)^{\alpha} = \sum_{k=1}^{\infty} {\alpha\choose{k}} x^k$
for $|x|<1$ the function $f(x)=(1+x)^{1/2}$ is defined on $A$ if the eigenvalues of $A$ satisfy $|\lambda_i| < 1 $ besides that $f(A)^2 = I + A$. This definitly works for $A=N$
The strange sum is this one $(1+x)^{\alpha} = \sum_{k=1}^{\infty} {\alpha\choose{k}} x^k$ I have never seen this one before. It looks familiar to binomial expansion. I can not understand where this sume comes from? 
Usually it is $(1+x)^{n} = \sum_{k=1}^{n} {n\choose{k}} x^k$
Where does this sum come from? What happends when $k>\alpha$?
Keep in mind that I have not studied integration theory.

There are several ways to think about this sum. The easiest (I think) is to say that we are simply using the Taylor series for the function $\sqrt{1 + x}$ centered at $x = 0$.  
In general, $\binom{\alpha}k$ is the coefficient of $x^k$ in the power-series expansion of $(1+x)^\alpha$. For an actual formula, see this page.
