Let $\mu_{D}(U)$ be the Haar measure on the D-dimensional Unitary group $U(D)$, where $U \in \mathrm{SU}(D)$ or $U(D)$. Can we think of this measure as picking first a unit vector according to the vector Haar measure in $D$ dimensions, then picking the second unit vector in the orthogonal subspace according to the vector Haar measure on $D-1$ dimensions, and so on? Namely, can the unitary Haar measure be decomposed as the product of unit vector Haar measures from 1 to D dimensions? If yes, how is this written formally?
I have looked into QR decomposition but it kind of does things the other way round. Essentially, it uses the Gram-Schmidt process where at first a vector is chosen then it is made orthogonal to the next chosen vector.
As pointed out by people in the comments QR decomposition indeed answers my question. The exact claim is actually proven in this paper (arXiv preprint here) at the beginning of page 5. I will write the proof below for reference.
The construction says that we can choose the unit vector $u_i$ (the $i$-th column of $U = \left(u_1, u_2, ...\right)$) recursively according to the unit vector Haar measure onto the orthogonal subspace of the previously chose unit vectors $u_1$, $...$, $u_{i-1}$. To prove that this is a unitary Haar measure we need to prove that the distribution of all the vectors is invariant under the action of any fixed unitary $V$, namely that $V u_1$, $V u_2$, $...$ are distributed identically to $u_1$, $u_2$, $...$. We will use $\sim$ to denote identically distributed variables, namely we will use $V u_i \sim u_i$ to denote that $V u_i$ and $u_i$ are distributed accordingly to the same distribution.
With these preliminaries in place let us start the proof. We will follow the proof of the paper and prove invariance only for $u_1$ and $u_2$ and them simply say that the invariance for $u_i$ follows in a similar manner (essentially we are leaving the proof by induction for the reader).
Let V be a fixed unitary. First notice that obviously $V u_1, V u_2 \cdots ,$ are mutually orthogonal (they are the columns of the unitary $VU$).
$u_1$. The first vector is trivial: since $u_1$ is a Haar random vector its distribution is left invariant. Hence, $V u_1$ has the same distribution as $u_1$.
Let $V_1$ be any unitary that leaves $V u_1$ unchanged, namely such that : \begin{align} V_1 V u_1 &= V u_1 \\ V^{-1} V_1 V u_1 &= u_1 \end{align} This means $u_1$ is an invariant subspace of $V^{-1} V_1V$ . Recall $u_2$ is chosen uniformly from the subspace orthogonal to $u_1$ and by left invariance of distribution of $u_2$, we have : \begin{align} V^{-1} V_1 V u_2 & \sim u_2 \\ V_1 V u_2 & \sim V u_2 \end{align} Since, $V_{1}$ on the orthogonal subspace of $V u_1$ is arbitrary this means that the distribution of $V u_{2}$ is also left invariant implying Haar distribution. This means that $V u_1$ and $V u_2$ are constructed with the same procedure as $u_1$ and $u_2$, meaning their distribution is the same and thus invariant under $V$.
This same argument can be repeated for the other vectors $u_3, ...$
This concludes the proof.