Proving zero raised to the zeroth power is equal to one using the power series? Its been a while since I have done math, so I wondering if there was a proof for this, if so what?

Is zero to the zero power 1? Or is it not defined?.

Thank you
Limits don't apply to 0^0 any more than they do to 1^oo. Consider

lim(n-&gt;oo) (1+1/n)^n = e

lim(n-&gt;oo) (1+2/n)^n = e^2


Both are of the form 1^oo, but neither of these equals 1. And surely 1^oo = 1. 

Also, various proposed limits for 0^0 come up with different answers:

lim(x-&gt;0) 0^x = 0

lim(x-&gt;0) x^0 = 1


So limits aren't helpful. But here's something that is helpful. 

Consider

e^x = SUM(n=0-&gt;oo) x^n/n!


We are told that this is good for all real numbers, no? It's even good for complex numbers. Now, 0 is a real number. So this formula is good for e^0. But that only works if 0^0=1. (Oddly, the same books that tell us that 0^0 is "undefined" also tell us that the e^x formula above is good for all real numbers.)

Therefore, if you insist on calling 0^0 "undefined", you have to make an exception for using the above formula to calculate e^0. I doubt anyone is willing to do so! So 0^0=1. 

This probably holds for any series in which you have the term a_0*x^0. We always equate this term as a_0, no matter what x is.

You get the same thing with the binomial theorem:

(a+b)^n = a^n + n * a^(n-1) * b + ... + b^n


In summation form you get

(a+b)^n = SUM(k=0-&gt;n) (n k) a^(n-k) b^k 

        = SUM(k=0-&gt;n) n!/(k!*(n-k)!) a^(n-k) b^k 

        = a^n b^0 + ... + a^0 b^n

        = a^n + ... + b^n


The summation version works because a^0 = 1 and b^0 = 1. Set a or b equal to zero and you get 0^0=1. And this is assuming n is a positive integer, where everyone agrees the theorem is true!

If you want to make the binomial theorem good for n=0, then 0^0=1 will do the job:

(a+b)^0 = 0!/(0!*0!) a^0 b^0 = a^0 b^0 = 1

0^0 = [a+(-a)]^0 = a^0 (-a)^0 = 1


No one has a problem defining 0! = 1. My high school book Modern Introductory Analysis additionally defines (n 0) = 1. So why not define (or conclude!) 0^0=1? 

Some say you can't determine 0^0 because a^0 = a^1 / a^1 and with a=0 you can't do the division. But this need not be the defining condition! You can determine it otherhow, as I have above. The exponent rule used for this is 

a^m * a^n = a^(m+n)


And a=0 satisfies it for integers m,n>=0. It satisfies the other exponent laws for non-negative exponents, too. 

In conclusion, 0^0 = 1 unless you make x = 0 an exception to 

e^x = SUM(n=0-&gt;oo) x^n/n!


and I really doubt anyone would.
One way you can arrive at this equality is by evaluating $$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{\ln(x^x)}  \\ = \lim_{x \to 0^+} e^{x\ln(x)}$$ Now using L'Hospital's rule on $x\ln(x) = \frac{ln(x)}{x^{-1}}$ we get $$\lim_{x\to 0^+} \frac{ln(x)}{x^{-1}} = \lim_{x\to 0^+} \frac{\frac{1}{x}}{-x^{-2}} \\ =\lim_{x\to 0^+} \frac{-x^2}{x} \\ = \lim_{x\to 0^+} =-x = 0$$ Hence $$\lim_{x \to 0^+} e^{x\ln(x)} = e^0 = 1$$ But the whole $0^0=1$ thing is pretty iffy. The right hand limit does not exist, for one thing.