Let $M =\begin{pmatrix} 1 & 0 & ... & 0 \\ 0 & 1 & ...& 0 \\ ... & ... & ... & ...\\ 0 & 0 & 0 & 1 \\ x_1 & x_2 &... & x_n \end{pmatrix} = \begin{pmatrix} &E & \\ x_1 & ... & x_n\end{pmatrix} \in \mathbb R^{(n+1) \times n}$
where $E \in \mathbb R^{n \times n} $ is the identity matrix.
Is there any easy way to show that $\det(M^T M) = 1 + x_1^2 + x_2^2 + ... + x_n^2 ? $
It should be obvious that $M^TM = E + K$, where $K$ is a symmetric matrix with $k_{ij} = x_ix_j$ for $1 \le i,j \le n $
Usually I would apply $\det (AB) = \det(A) \det(B) $, but that doesn't work because $M \in \mathbb R^{(n+1) \times n}$.
Anyone got a hint?
As you noticed $$M^TM = E + K,$$ where $(K)_{ij}=k_{ij} = x_ix_j$, that is ${\rm rank}(K)=1$.
In this case, there is a general fact, that if ${\rm rank}(A)=1$, then $$\det(E+A) = 1+{\rm tr}(A).$$ It can be proved easily by considering the normal Jordan form of a matrix $A$. Since the rank of $A$ is 1, then the Jordan form of a $A$ has simple form.
By this it follows $${\rm det}(M^TM) = 1 + {\rm tr}(K) = 1+x_1^2+\cdots+x_n^2.$$