Both $0^0$ and $0^1$ are undefined? Following up on my previous question at What values of $0^0$ would be consistent with the Laws of Exponents? I am still trying to get a handle on $0^0$. It now seems to me that not only is it undefined, but so is $0^1$. What mistake, if any, am I making here? (Yes, I know, you can simply define values for both and hope no contradictions arise, but I don't find such an approach to be very satisfying.)
In the natural numbers, exponents greater than 1 are naturally defined as follows:
$x^2=xx$
$x^3=xxx=x^2x$
$x^4=xxxx=x^3x$
$x^5=xxxxx=x^4x$
and so on.
Therefore $x^{n+1}=x^nx$ for $n\gt 1$.
What about $x^0$ and $x^1$?
Suppose $x^1x=x^2$. Then $x^1=x^2/x = x$ for $x\ne 0$.
Suppose $x^0x=x^1$. Then $x^0=x^1/x = 1$ for $x\ne 0$.
Thus $x^0$ and $x^1$ are both undefined for $x\ne 0$.
CONCLUSION:
For proof that $0^1=0$ assuming the usual Laws of Exponents, see comments by Henry Swanson below. 
EDIT:
See FOLLOW-UP in my answer below.
 A: In set theory and combinatorics, and for purposes of expansions in power series, $0^0$ is $1$.
It is $1$ because the number of factors multiplied is $0$.  When you multiply no numbers, you get $1$, just as when you add no numbers at all you get $0$.  Not multiplying by anything is the same as multiplying by $1$, just as adding no numbers at all is the same as adding $0$.  Google the term "empty product" and read about it.
"Everybody knows" that
$$
e^z = \sum_{n=0}^\infty \frac{z^n}{n!}.
$$
But when $z=0$, then the first term is $\dfrac{0^0}{0!}$.  But the first term must be $1$.
One the other hand $0^0$ is an indeterminate form in that if $f$ and $g$ approach $0$ as $x$ approaches something, then $f^g$ could approach any positive number at all or $0$ or $\infty$, depending on which functions $f$ and $g$ are.  (But if $(f,g)$ approaches $(0,0)$ from within a sector bounded by two lines of positive slope, then the limit is $1$.)
A: Yes, for $x \ne 0$, $x^1 = x^2 / x$. Does that mean $0^1$ is undefined? No. We also know that for $x \ne 0$, $x^n = x^{n+1} / x$. So is $0^n$ undefined? No.
However, we must take care to define $0^1$ consistently. If you look at the function $x^y$ (or, if you don't know how to define exponents for real numbers, $e^{y \ln x}$) at the point $(0,1)$, you will see it behaves nicely.
But at $(0,0)$, it doesn't, so many people leave it undefined.
A: $0^0$ can be defined as $1$ for the following reason:  


*

*The alternating sum of binomial coefficients from the $n$-th row of Pascal's triangle is what you obtain by expanding $(1-1)^n$ using the binomial theorem, i.e., $0^n$. But the alternating sum of the entries of every row except the top row is $0$, since $0^k=0$, $\forall k > 1$. But the top row of Pascal's triangle contains a single $1$, so its alternating sum is $1$, which supports the notion that $(1-1)^0=0^0$, if it were defined, should be 1.

A: You can't conclude that $x^1$ is undefined for $x \neq 0$ just because division by $0$ is undefined. The reason that $x^0$ is undefined for $x=0$ is rather more subtle. Moreover, I'm failing to see why we can't conclude that $x^n$ is undefined for $x=0$ by your logic? Can't we say $\displaystyle x^n = \frac{x^{n+1}}{x}$ for $x \neq 0$ ?
To know that why $0^0$ is undefined you can look at the behavior of the function$x^y$ near $(0,0)$ to see that $x^y$ has serious complications around $(0,0)$. That's why $x^y$ is left undefined for $(x,y)=(0,0)$, because we can't define a particular value for it. It's like that you define a binary operation $x*y$ by $\displaystyle x*y=\frac{\sin(x)}{y}$ and you insist on defining $0*0$.
See also here on wikipedia. Cauchy was the first one who objected to $0^0=1$.
A: $$x^{n-m}=\underbrace{x^{-1}\ldots\cdot x^{-1}\cdot x^{-1}}_\text{m times} \cdot 1\cdot  \underbrace{x\ldots\cdot x\cdot x}_\text{n times}$$
A: $0^1$ is surely defined, and it's zero (0)
as for $0^0$, it is undefined, becuase if x=0 we'll get:
$0^{x-1} = 0^{-1} = \frac{1}{0}$ which is really undefined expression.
BUT, there is a convention I read about over the net that $0^0 = 1$, without further reasoning
A: 
In the natural numbers, exponents greater than 1 are naturally defined as follows:
$x^2=xx$
$x^3=xxx=x^2x$
$x^4=xxxx=x^3x$
$x^5=xxxxx=x^4x$
and so on.

More formally, exponentiation can be defined as a binary function on $N$ such that
(1) $\forall x\in N:x^2=xx$
(2) $\forall x,y\in N:x^{y+1}=x^y x$
It can then be shown that
(1) $\forall x\in N:(x\ne 0 \implies x^1=x)$
(2) $\forall x\in N:(x\ne 0 \implies x^0=1)$
(3) $\forall x,y,z\in N:(x^{y+z}=x^y x^z) \implies 0^1=0 \land (0^0=0 \lor 0^0=1)$
Formal proof in DC Proof format (126 lines) at http://dcproof.com/Ambiguity-of-0-to-the-0.htm
FOLLOW-UP
Down vote quite justified! My thinking on this topic has evolved considerably in the weeks since I posted the above, although I still have exponentiation being defined as a binary function on $N$ such that
(1) $\forall x\in N: (x\ne 0 \implies x^0=1)$, or equivalently $\forall x\in N:x^2=xx$
(2) $\forall x,y\in N:x^{y+1}=x^y x$
$0^0$ is assumed to be a natural number, but no is actual value is assigned to it. Contrary to my previous claim, however, it can be shown that we must have $0^1=0.$
See details and formal proofs at "Oh, the Ambiguity!" at my math blog.
