Is there a way to prove this statement about vector spaces. For finite dimensional vector spaces over a field with q elements: The number of $k$ dimensional subspaces of an $n$ dimensional vector space, $[n,k]$ is equal to the number of $k-1$ dimensional subspaces of an $n-1$ dimensional vector space, plus $q^k$ times the number of $k$ dimensional subspaces of an $n-1$ dimensional vector space. In other words, $[n,k] = [n-1,k-1] + q^k*[n-1,k]$.
It appears that the number of times that any particular vector appears in a $k$ dimensional subspace of an $n$ dimensional space is equal to $[n-1,k-1]$ while the number of subspaces missing this particular vector is $q^k*[n-1,k].$
Is there a way to prove this statement which I am making based on staring at subspaces of small dimensions.
 A: The reasoning in your second paragraph is correct. To formalize, let $V$ be a $n$ dimensional vector space over the finite field with $q$ elements and $v$ be an arbitrary (non-zero) vector of $V$. Let
$$\begin{align}A&=\{E \mid \text{$E$ subspace of $V$ such that} \dim{E}=k \text{ and } v \in E\}, \\
B&=\{E \mid \text{$E$ subspace of $V$ such that} \dim{E}=k \text{ and } v \notin E\}.
\end{align}
$$
Let $H$ be an arbitrary complementary subspace of $\operatorname{Span}(v)$ in $V$, and $$A'=\{F \mid \text{$F$ subspace of $H$ such that} \dim{F}=k-1\}.$$
Since $\dim{H}=n-1$, $A'$ contains exactly $[n-1,k-1]$ elements. But the function $A' \to A$, that maps $F$ to $F \bigoplus \operatorname{Span}(v)$ is a bijection (its inverse is given by $E \mapsto E\cap H$). Hence $A$ and $A'$ have the same cardinality.
The computation for $B$ seems a little more tricky. Let $\varphi(k)$ be the number of basis of a given $k$-dimensional subspace (we will not need to compute $\varphi(k)$), and
$$
\begin{align}
B_1&=\{(F,f_1,\dotsc,f_k) \mid \text{$F$ subspace of $H$ such that} \dim{F}=k, (f_1,\dots,f_k) \text{ basis of $F$}\}, \\
B_2&=\{(E,e_1,\dotsc,e_k) \mid \text{$E$ subspace of $V$ such that} \dim{E}=k, (e_1,\dots,e_k) \text{ basis of $E$}, v\notin E\}.
\end{align}
$$
Then $\operatorname{Card}{B_1}=[n-1,k]\times\varphi(k)$ and $\operatorname{Card}{B_2}=\operatorname{Card}{B}\times\varphi(k)$. We now consider the function $B_1 \times \mathbb{F}_q^k \to B_2$, that maps
$$(F,(f_1,\dotsc,f_k),(\lambda_1,\dots,\lambda_k))
$$
to
$$(E=\operatorname{Span}(f_1+\lambda_1v,\dots,f_k+\lambda_kv),(f_1+\lambda_1v,\dots,f_k+\lambda_kv)).
$$
This is a bijection : one can explicitly write its inverse using the projection along $\operatorname{Span}(v)$ onto $H$. So $\operatorname{Card}{B_1} \times q^k=\operatorname{Card}{B_2}$, which yields $[n-1,k]\times q^k=\operatorname{Card}{B}$.
To summarize
$$
\begin{align}
[n,k]&=\operatorname{Card}{A}+\operatorname{Card}{B} \\
&=[n-1,k-1]+[n-1,k]\times q^k.
\end{align}
$$
