Number of $n-1$-dimensional subspaces of $n$-dimensional space over finite field

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I got a question with two parts.

Let $V$ be a $n$-dimensional vector space over $\mathbb{F}_{p}$ - finite field with $p$ elements.

a) How many $1$-dimensional subspaces $V$ has.

b) How many $n-1$-dimensional subspaces $V$ has.

I solved (a) with action of the multiplicative group of $\mathbb{F}_{p}$ on $V$, but I didn't succeed to solve (b) with similar idea..

I still prefer an idea with action of groups.. thanks !

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Hint:

There is a bijection between subspaces of dimension $k$ and $k\times n$ matrices of rank $k$ in reduced row-echelon form.

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This answer will only be helpful if you have some background in (projective) geometry.

The number of $1$-dimensional subspaces in a vector space of dimension $n$ over the field $\mathbb F_q$ corresponds to the number of points in a projective space of (geometric) dimension $n-1$ over the same field, denoted by $|\mathbb P^{n-1}(\mathbb F_q)|$. Since a projective space can be thought of as the disjoint union of an affine space of the same dimension and a hyperplane at infinity, i.e. a projective space of a dimension one smaller, we have that

$$ |\mathbb P^{n-1}(\mathbb F_q)| = |\mathbb A^{n-1}(\mathbb F_q)| + |\mathbb P^{n-2}(\mathbb F_q)| = q^{n-1} + |\mathbb P^{n-2}(\mathbb F_q)| . $$

Therefore, inductively, the answer is given by:

$$ |\mathbb P^{n-1}(\mathbb F_q)| = q^{n-1} + q^{n-2} + \dots + p + 1 = \frac{q^n-1}{q-1}. $$

Because of projective duality, a projective space has the same number of points and hyperplanes so this is actually the answer to the other question as well.