n^n

We know o^n = 0 for all n <> 0

We also know n^0 = 1 for all n <> 0

But why is n^n undefined for n = 0 but in the limit n^n = 1?

I thought the limit didn't exist

It does.

Consider the limit as x->0+ x^x = lim x->0+ e^(xlogx)

I know the answer, lim n->0 (n^n) = 1 but just don't know why 0^0 is undefined.

I suppose it's so you can't do crazy stuff like $1=0^0=0^{1\cdot 0}=0^10^0=0\cdot 1 = 0$

I know the answer, lim n->0 (n^n) = 1 but just don't know why 0^0 is undefined.

That's the limit from above. I just mean the limit.

If you do the series expansion of e^(xlogx) and then let x->0, you'll be left with 1.

Surely your 0^0 = 0^(1.0) doesn't make sense?

OK, I'll take your word for it. Infinite series don't always converge so I guess one would have to deal with that issue

I suppose it's so you can't do crazy stuff like $1=0^0=0^{1\cdot 0}=0^10^0=0\cdot 1 = 0$

Since when could you do a^(b*c)=(a^b)*(a^c)?

If I'm doing set theory then I argue that $0^0=1$, since we can define $a^b$ to be the cardinality of the set of functions $B \to A$ where $\left| A \right| = a$ and $\left| B \right| = b$. When $A=B=\varnothing$, this set consists only of the empty function, so its cardinality is 1.

If I'm doing analysis then I argue that $0^0$ is undefined, since for instance $\lim_{x \to 0} x^0 = 1$ but $\lim_{x \to 0} 0^{\left|x\right|} = 0$.