= Solution
The code $\operatorname{RM}(d,0)$ is the binary repetition code of length $2^d$, while
$$
\operatorname{RM}(d,d)=\mathbb F_2^{\,2^d}
$$
is the full binary code. For $0<r<d$, the <Reed-Muller bar-product recursion> is
$$
\operatorname{RM}(d,r)
=\operatorname{RM}(d-1,r)
\mid\operatorname{RM}(d-1,r-1).
$$
Thus its rank $k(d,r)$ satisfies Pascal's recursion
$$
k(d,r)=k(d-1,r)+k(d-1,r-1),
$$
with the stated boundary values. Therefore
$$
\boxed{k(d,r)=\sum_{j=0}^r\binom dj}.
$$
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