All tangential Stiefel-Whitney numbers of $\mathbb{R}P^q$ are zero iff $q$ is odd, reference? Where can I find a reference to the proof of the following fact?

All tangential Stiefel-Whitney numbers of $\mathbb{R}P^q$ are zero if and only if $q$ is odd.

I made a quick search through Hatcher but could not find any mention...
 A: I don't know of a reference of this fact, but it follows from a collection of well-known results.
First of all, the total Stiefel-Whitney class of $T\mathbb{RP}^q$ is $w(T\mathbb{RP}^q) = (1 + a)^{q+1}$, so $w_i(T\mathbb{RP}^q) = \binom{q+1}{i}a^i$ (as this is an element of $H^i(X, \mathbb{Z}/2\mathbb{Z})$, reduce $\binom{q+1}{i}$ modulo $2$). 
If $q$ is even, $w_q(T\mathbb{RP}^q) = \binom{q+1}{q}a^q = (q+1)a = a$, so the corresponding Stiefel-Whitney number is $w_q = \langle w_q(T\mathbb{RP}^q), [\mathbb{RP}^q]\rangle = \langle a^q, [\mathbb{RP}^q]\rangle = 1 \neq 0$. More generally, the top Stiefel-Whitney class is the mod $2$ reduction of the Euler class so the corresponding Stiefel-Whitney number is the Euler characteristic modulo $2$ (in this case $\chi(\mathbb{RP}^q) = 1$). 
So the only possible values of $q$ for which all the Stiefel-Whitney numbers of $\mathbb{RP}^q$ can vanish are the odd ones. We still need to show that for all odd values of $q$, the numbers do indeed vanish.
For any $i_1, \dots, i_k$ with $i_1 + \dots + i_k = q$, we have 
\begin{align*}
w_{i_1,\dots,i_k} &= \langle w_{i_1}(T\mathbb{RP}^q)\cup\dots\cup w_{i_k}(T\mathbb{RP}^q), [\mathbb{RP}^q]\rangle\\ 
&= \left\langle\binom{q+1}{i_1}a^{i_1}\cup\dots\cup\binom{q+1}{i_k}a^{i_k}, [\mathbb{RP}^q]\right\rangle\\
&= \binom{q+1}{i_1}\dots\binom{q+1}{i_k}\langle a^q, [\mathbb{RP}^q]\rangle\\ 
&= \binom{q+1}{i_1}\dots\binom{q+1}{i_k}.
\end{align*}
This is zero if and only if one of the binomial coefficients are even. 
To show the existence of some even binomial coefficient in the above product, I need to use the following result: if $n$ is even and $k$ is odd, $\binom{n}{k}$ is even. This follows from Lucas' Theorem which is a useful thing to know when computing Stiefel-Whitney classes/numbers, but there is also a simpler proof. First we start with the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$. As $n$ is even, the right-hand side of the identity is divisible by $2$, and therefore so is the left-hand side. As $k$ is odd, $\binom{n}{k}$ must be divisible by $2$, that is, even.
Returning to the problem at hand, note that if $i_1 + \dots + i_k = q$ and $q$ is odd, there is some $1 \leq j \leq k$ such that $i_j$ is odd. By the above result $\binom{q+1}{i_j}$ is even and hence $w_{i_1,\dots,i_k} = 0$.
So we have established that all the Stiefel-Whitney numbers of $\mathbb{RP}^q$ vanish if and only if $q$ is odd. 

By a theorem of Thom and Pontryagin, two closed manifolds are cobordant if and only if they have the same Stiefel-Whitney numbers. In particular, a closed manifold has all Stiefel-Whitney numbers zero if and only if it is the boundary of some compact manifold (i.e. null-cobordant). So, for every odd $q$, there is some compact $(q+1)$-dimensional manifold $X_{q+1}$ with $\partial X_{q+1} = \mathbb{RP}^q$. The manifolds $X_{q+1}$ can actually be determined, see this MathOverflow question.

Added Later: I just realised that the solution I gave above is a less elegant version of the solution outlined in Milnor & Stasheff on page $51$. It may not be suitable as a reference for the highlighted fact in your post though. I have included the relevant excerpt below.

As an example, let us try to compute the Stiefel-Whitney numbers of the projective space $P^n$ (which is about the only manifold we are able to handle at this point). Let $\tau$ denote the tangent bundle of $P^n$. If $n$ is even, then the cohomology class $w_n(\tau) = (n+1)a^n$ is non-zero, and it follows that the Stiefel-Whitney number $w_n[P^n]$ is non-zero. Similarly, since $w_1(\tau) = (n+1)a \neq 0$, it follows that $w_1^n[P^n]\neq 0$. If $n$ is actually a power of $2$, then $w(\tau) = 1 + a + a^n$, and it follows that all other Stiefel-Whitney numbers of $P^n$ are zero. In any case, even if $n$ is not a power of $2$, the remaining Stiefel-Whitney numbers can certainly be computed effectively as products of binomial coefficients.
On the other hand if $n$ is odd, say $n = 2k - 1$, then $w(\tau) = (1 + a)^{2k} = (1+a^2)^k$, so it follows that $w_j(\tau) = 0$ whenever $j$ is odd. Since every monomial of total dimension $2k - 1$ must contain a factor $w_j$ of odd dimension, it follows that all of the Stiefel-Whitney numbers of $P^{2k-1}$ are zero. This gives some indication of how much detail and structure this invariant overlooks.

