Wreath product is associative but regular wreath product is not associative. I am currently reading An Introduction of Theory of Groups by Rotman. In page 174, Theorem 7.26 states that wreath product is associative. But in the next page, regular wreath product $W=D \wr_r Q$ is defined when $Q$-set acting on itself by left multiplication. Note that $|D\wr_r Q|=|D|^{|Q|}|Q|$. So regular wreath product is not associative when all groups are finite because $$|T \wr_r (D \wr_r Q)|\neq |(T \wr_r D) \wr_r Q|$$ But if regular wreath product is a special case of wreath product, the operation should be also associative.

Wreath product is associative in the sense that, if $\Omega$ is a $Q$-set and $\Lambda$ is a $D$-set, $$T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)\simeq (T\wr_\Lambda D)\wr_{\Omega} Q,$$ for there is a natural action of $D\wr_\Omega Q$ on $\Lambda\times\Omega$ induced by the ones of $D,Q$ on $\Lambda,\Omega$ respectively, and showing the isomorphism is just a matter of making definitions explicit.

Note that in this case you have, if all groups and actions are finite, $$|T\wr_{\Lambda\times\Omega}(D\wr_\Omega Q)|=|T|^{|\Lambda||\Omega|}|D|^{|\Omega|}|Q|=(|T|^{|\Lambda|}|D|)^{|\Omega|}|Q|=|(T\wr_\Lambda D)\wr_{\Omega} Q|.$$

Now, if the product on the right is regular (that is to say, $\Lambda=D$ and $\Omega=Q$ with the natural actions), you still get an isomorphism $T\wr_{D\times Q}(D\wr_Q Q)\simeq (T\wr_D)\wr_Q Q$. Thing is, though, that the outer product on the left is not regular, since the sets $D\wr_Q Q=\left(\prod_{q\in Q} D\right)\times Q$ and $D\times Q$ differ in cardinality.