Existence (by induction on $n$). For $n = 1$: $A = (a_{11})$, $L = (1)$, $U = (a_{11})$.

For the inductive step, partition $A = \begin{pmatrix} A_{n-1} & c \\ r^T & a_{nn} \end{pmatrix}$ where $A_{n-1}$ is $(n-1) \times (n-1)$. By hypothesis, $\det(A_{n-1}) \neq 0$ (it is a leading principal minor), so by induction $A_{n-1} = L_{n-1} U_{n-1}$.

Seek $L = \begin{pmatrix} L_{n-1} & 0 \\ \ell^T & 1 \end{pmatrix}$, $U = \begin{pmatrix} U_{n-1} & u \\ 0 & u_{nn} \end{pmatrix}$.

Then $LU = \begin{pmatrix} L_{n-1}U_{n-1} & L_{n-1}u \\ \ell^T U_{n-1} & \ell^T u + u_{nn} \end{pmatrix} = \begin{pmatrix} A_{n-1} & c \\ r^T & a_{nn} \end{pmatrix}$.

This gives: $L_{n-1} u = c$, so $u = L_{n-1}^{-1} c$ (solvable since $L_{n-1}$ is nonsingular); $\ell^T U_{n-1} = r^T$, so $\ell^T = r^T U_{n-1}^{-1}$ (solvable since $U_{n-1}$ is nonsingular, as $\det(U_{n-1}) = \det(A_{n-1})/\det(L_{n-1}) = \det(A_{n-1}) \neq 0$); and $u_{nn} = a_{nn} - \ell^T u$.

Uniqueness. Suppose $A = L_1 U_1 = L_2 U_2$. Then $L_2^{-1} L_1 = U_2 U_1^{-1}$. The left side is unit lower triangular and the right side is upper triangular. A matrix that is simultaneously unit lower and upper triangular must be $I$. Hence $L_1 = L_2$ and $U_1 = U_2$.

Necessity. If $A = LU$ exists, then $A[1:k, 1:k] = L[1:k, 1:k] \cdot U[1:k, 1:k]$, so $\det(A[1:k, 1:k]) = \det(L[1:k, 1:k]) \cdot \det(U[1:k, 1:k]) = \prod_{i=1}^k u_{ii}$. For the factorization to succeed, we need each $u_{kk} \neq 0$ during elimination, which requires $\det(A[1:k,1:k]) \neq 0$.

Pivoted factorization. If $A$ is nonsingular but has a zero leading principal minor, there exists a row permutation $P$ (constructed by partial pivoting during Gaussian elimination) such that all leading principal minors of $PA$ are nonzero. $\square$