Determinant of a matrix of the form $M = I + xx^t$ $$M=\begin{pmatrix} 1+x_1^2 & x_1x_2 &...&x_1x_n \\ x_2x_1 & 1+x_2^2 &...&x_2x_n \\...&...& &...& \\x_nx_1 & x_nx_2& ...&1+x_n^2&\end{pmatrix}.$$
So I noticed that $M$ is a symmetric matrix and $ M=I+(x_1, x_2,...,x_n)^T(x_1, x_2,...,x_n)$. That's all I can get.
 A: You can continue from your observation using the matrix-determinant lemma (which is not hard to prove, by the way).
For a column vector $x = (x_1,...,x_n)$, and an $n\times n $ identity matrix $I$, we have
$$
M = I + x x^T 
$$
hence, in view of the aforementioned lemma, 
$$
\det M = (1 + x^T I x)\det I = 1 + \|x\|^2 = 1 + x_1^2 + \cdots +x_n^2.
$$
A: Consider the matrix
$$N=M-I.$$
It has rank $\le1$, so its characteristic polynomial $P(\lambda)=\det(\lambda I-N)$
has the form $\lambda^n-a\lambda^{n-1}$. But $a$ is the trace of $N$, so
$$P(\lambda)=\lambda^n-(x_1^2+\cdots+x_n^2)\lambda^{n-1}.$$
Now
$$\det M=\det(I+N)=(-1)^nP(-1)=1+x_1^2+\cdots+x_n^2.$$
A: Denote by $M_n$ the matrix we want to compute the determinant. Letting $\mathbf{x}=(x_1,\dots,x_{n-1})$, we express $M_n$ as a block matrix:
$$
M_n=\pmatrix{M_{n-1}& x_n\mathbf x^t\\
x_n\mathbf{x}&1+x_n^2}.
$$
Since $\pmatrix{  x_n\mathbf x^t\\1+x_n^2}=\pmatrix{  0\\1}+\pmatrix{  \mathbf x_nx^t\\x_n^2}$, it follows by multi-linearity of the determinant that 
$$
\det\left(M_n\right)=\det\pmatrix{M_{n-1}& 0\\
x_n\mathbf{x}&1}+\det\pmatrix{M_{n-1}& x_n\mathbf x^t\\
x_n\mathbf{x}&x_n^2}.
$$
The first term of the right hand side is $\det\left(M_{n-1}\right)$; for the second one, using two times the multi-lineartiy of the determinant (with respect to lines and then columns), we derive that 
$$
\det\left(M_{n}\right)=\det\left(M_{n-1}\right)+x_n^2\det\pmatrix{M_{n-1}&  \mathbf x^t\\
 \mathbf{x}&1}.
$$
The last determinant is one: one can make the linear combinations $C_i\leftarrow x_iC_n$ to see this.
