How to show that $\det(S+dd^T)=\det S\cdot(1+d^TS^{-1}d)$?

Let $x_i,\overline x, d$ (with $i\leq n$) be given vectors (column, of the same length). Put $S=\frac{1}{n}\sum_{i=1}^n(x_i-\overline x)(x_i-\overline x)^T$. Assume that $S$ is invertible. Is the following identity true, and if yes, how can I show it? $$\det(S+dd^T)=\det S\cdot(1+d^TS^{-1}d)$$

Thoughts: obviously, we first take the $S$ out of the determinant to obtain $\det(I+dd^TS^{-1})$, then we can take a transpose to get $\det(I+S^{-T}d^Td)$, which is of course equal to $\det(I+S^{-1}d^Td)$, which looks almost, but not quite right...

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Note that $\mathrm{det}(I+xx^T)=1+x^Tx$ since $xx^T$ is a rank one matrix with eigenvector $x/\|x\|_2^2$ and eigenvalue $\|x\|_2^2=x^Tx$. Therefore $x$ is an eigenvector of $I+xx^T$ with eigenvalue $1+x^Tx$. Since all other eigenvalues are 1, this is also the determinant. Now use $S+dd^T=S^{1/2}\left(I+(S^{-1/2}d)(S^{-1/2}d)^T\right)S^{1/2}$.
In your case, $S$ is a sum of positive semi-definite matrices, so it itself is positive semi-definite, and therefore a unnique positive semi-definite square-root exists. – Peder Jan 27 '13 at 20:50