Another basic Ordinal arithmetic question: $\omega \cdot (\omega+1)=?$

I am getting confused about ordinal arithmetic
I am trying to figure out what $\omega \cdot (\omega+1)$ is.
The fact that the distributive law applies from the left side makes me think that the answer is $\omega^2 + \omega$. However applying the supremum definition of multiplication I am getting a different answer. The supremum method is giving me: $$\omega \cdot (\omega+1) = \sup \{\omega \cdot n \ | \ n \in \omega+1 \} = \omega^2$$ Since $\omega^2 \geq \omega \cdot n, \forall n < \omega + 1$.

Which way is correct and why is the other method wrong here?

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$\omega+1$ is not a limit. It is a successor. –  Chris Eagle Jan 16 '13 at 21:33
$\omega^2+\omega$ is correct: you have a string of $\omega+1$ copies of $\omega$. @Chris has identified the error in the other approach. –  Brian M. Scott Jan 16 '13 at 21:34

You are wrong in the definition of multiplication. It holds that $\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma$, but $(\alpha+\beta)\cdot\gamma\neq\alpha\cdot\gamma+\beta\cdot\gamma$ in some of the cases.
It follows that $\omega\cdot(\omega+1)=\omega\cdot\omega+\omega\cdot1=\omega^2+\omega$.
The second method is wrong, because the definition of multiply that you used in second method is right only when the second ordinal is a limit ordinal, so if the second ordinal is something like $\omega+1$ the definition of multiply is the distributive law :$a(b+1) = ab+a$