Let $$ M =\begin{pmatrix} a & v^\top \\ v & B \end{pmatrix}, $$ with $a \in \mathbb{R}$ and $a \neq 0$, $v \in \mathbb{R}^{n-1}$, $B \in \mathbb{R}^{n-1 \times n-1}$ invertible and symmetric.
What is a simple expression for $(M^{-1})_{11}$, i.e., a nice simplification of the block inversion formula?
My suspect (coming from another calculation) would be that $$ (M^{-1})_{11} = \left(a - v^\top B^{-1} v\right)^{-1}, $$ but I cannot prove it.
EDIT:
In light of the accepted answer below, we have $$ (M^{-1})_{11} = \frac{\det(B)}{\det(M)}, $$ but, since $B$ is invertible, it holds (see here) $$ \det(M)= \det(B)\det(a- v^\top B^{-1}v)=\det(B)(a- v^\top B^{-1}v). $$ Hence $$ (M^{-1})_{11} = \frac{\det(B)}{\det(B)(a- v^\top B^{-1}v)} = (a- v^\top B^{-1}v)^{-1}, $$ my suspect.
You're can find a solution in the following formula for the inverse of the matrix $M$: $$M^{-1}=\frac{1}{\det M} C^T,$$ where $C$ is the cofactor matrix, i.e. the matrix with entries $C_{ij}=\det D^{ij}$ where $D^{ij}$ is the matrix obtained from $M$ by removing its $i$th row and $j$th column.
In your case, $D^{11}=B$, hence $$(M^{-1})_{11}=\frac{\det B}{\det M}.$$
Here's the wikipedia article on this: https://en.wikipedia.org/wiki/Minor_(linear_algebra)#Inverse_of_a_matrix