The exterior derivate I am trying to show that $d(a \wedge da)=0$ if $k$, the degree of k-form $a$ is even. I have said: $=da \wedge da + (-1)^k a \wedge d^2a$ I believe the first term is zero due to repeated indices and the second is zero since $d^2=0$. However, a subsequent exercise relies on the fact that $d(a \wedge da)$ is not alway zero if k is odd. Can anybody point out the error that I have made?

It is the second term that is always zero (because $d^2a = 0$), no matter what $k$ is. The first term is zero because $da$ has odd degree. In general

$$ x \wedge y = (-1)^{|x||y|} y \wedge x$$

where $|x|,|y|$ are the degrees of $x,y$. In particular, if $x$ has odd degree then $x \wedge x = 0$.