cardinal exponentiation, $k^{<\lambda}$ I have the following well-known exercise in cardinal arithmetics:
If $\kappa, \lambda$ are cardinals such that $\lambda$ is infinite, then $\kappa^{<\lambda}$ equals the supremum of the $\kappa^{\theta}$, where $\theta < \lambda$ is a cardinal.
1) Isn't this false for $\kappa=1$? The left side is $\lambda$ and the right side $1$. Remark that $\kappa^{<\lambda}$ is defined to be the cardinality of the disjoint(?) union of the function sets $map(\alpha,\kappa)$, where $\alpha < \lambda$ is an ordinal.
2) Is this supremum understood as a cardinal supremum or an ordinal supremum? If the latter is the case, why should the supremum be a cardinal at all?
3) Anyway, I don't know how to solve this exercise. I've already seen it several times, but 1) and 2) remain obscure in the literature I know.
 A: First, (2): if you define a cardinal to be an ordinal that is not bijectable with any strictly smaller ordinal, then suppose that $\alpha$ is the ordinal supremum of $\kappa^{\theta}$, $\theta\lt\lambda$. Let $\beta$ be a strictly smaller ordinal than $\alpha$. By the definition of supremum, there must exist a $\theta\lt\lambda$ such that $\beta\lt\kappa^{\theta}\leq\alpha$; in particular, $\kappa^{\theta}$ cannot be bijected with $\beta$ (being a cardinal), and therefore neither can $\alpha$ (that would give an embedding of $\kappa^{\theta}$ into $\beta$, and Cantor-Bernstein would give you $|\beta|=\kappa^{\theta}$, contradicting that the latter is not bijectable with any strictly smaller ordinal). Thus, $\alpha$ is not bijectable with any strictly smaller ordinal, and so must be a cardinal. So whether you define the supremum as the ordinal-sup or the cardinal-sup, you will still get a cardinal (this holds for any set of cardinals).
Second, (1): you are correct that the definition does not match this for $\kappa=1$ (or for $\kappa = 0$); as noted by Carl in the comments, this is likely an erratum or ommission; it should hold for any $\kappa>1$.
Edit: The definition as a sum is done running over all ordinals, rather than all cardinals, so I'm fixing this below.
Finally, (3): you are trying to show that the supremum of the $\kappa^{\theta}$ equals the sum over all ordinals $\alpha<\lambda$ of $|\kappa^{\alpha}$, assuming $\kappa\geq 2$ (to prevent the problems noted). This because $|\kappa^{\alpha}|=|map(\theta,\kappa)|$ by definition, and the cardinality of the disjoint union is the cardinal sum. I believe this can be shown by transfinite induction on $\lambda$ as I do below, but there probably is a simpler method.
So, the proposition we want to show is that for any infinite ordinal $\lambda$, we have
$$\sup_{\alpha\lt\lambda}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}|\kappa^{\alpha}|.$$
First, the equality holds for $\lambda=\omega$: if $2\leq \kappa\lt\aleph_0$, then $\sup\{|\kappa^n|\,|\, n=0,1,2,3,\ldots\} = \aleph_0$ and $\sum_{n=0}^{\infty}|\kappa^n| = \aleph_0$; if $\aleph_0\leq\kappa$, then $|\kappa^n|=\kappa$ for all $n$, and $\sum_{n=0}^{\infty}|\kappa^n| = \sum_{n=0}^{\infty}\kappa = \kappa\aleph_0=\kappa$, so both sides agree.
Assume the result holds for $\lambda$; then the supremum of $|\kappa^{\alpha}|$ with $\alpha\lt\lambda^+$ is $|\kappa^{\lambda}|$; on the other hand, 
$$\sum_{\alpha\lt\lambda^+}|\kappa^{\alpha}| = \left(\sum_{\alpha\lt\lambda}|\kappa^{\alpha}|\right) + \kappa^{\lambda} = \sup_{\alpha\lt\lambda}|\kappa^{\alpha}|+|\kappa^{\lambda}|=|\kappa^{\lambda}|,$$
where the last equality holds because $|\kappa^{\alpha}|\leq |\kappa^{\lambda}|$ for each $\alpha\lt \lambda$, so the supremum is at most $|\kappa^{\lambda}|$, and the sum of two infinite cardinals is equal to their maximum. So again the two expressions agree. 
Finally, we want to show that if $\lambda$ is a limit ordinal and the result holds for all $\beta\lt\lambda$, then it holds for $\lambda$. Then
$$\sup_{\alpha\lt\lambda}(|\kappa^{\alpha}|) = \sup_{\beta\lt\lambda}\left(\sup_{\alpha\lt\beta}(|\kappa^{\alpha}|\right) = \sup_{\beta\lt\lambda}\sum_{\alpha\lt\beta}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}(|\kappa^{\alpha}|).$$
So the equality holds for $\lambda$ as well. This establishes the result by transfinite induction for all infinite ordinals $\lambda$
Further Edit, 2 Sep 2010: Clarify how to finish it off.
So, the above shows that for any ordinal $\lambda$, $\sup_{\alpha\lt\lambda}|\kappa^{\alpha}| = \sum_{\alpha\lt\lambda}|\kappa^{\alpha}|$. To finish the exercise, we need to show that if $\lambda$ is a cardinal (that is, an ordinal that is not bijectable with any strictly smaller ordinal), then $\sup\{|\kappa^{\alpha}|\colon\alpha$ is an ordinal and $\alpha\lt\lambda\} = \sup\{|\kappa^{\theta}|\colon \theta$ is a cardinal and $\theta\lt\lambda\}$. To see this, note that $|\kappa^{\alpha}|=|\kappa|^{|\alpha|}$, and since $\lambda$ is assumed to be a cardinal, if $\alpha\lt\lambda$, then there exists a cardinal $\theta$, $\theta\lt\lambda$, such that $|\alpha|=|\theta|$, and hence $|\kappa^{\alpha}|=|\kappa^{\theta}|$.  Thus, the two sets are equal, so their suprema are equal as well. 
