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.
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 $1$.)
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 &quot;Oh, the Ambiguity!&quot; at my math blog.