A classical way is to use the induction. Let $A\in\mathbb{R}^{n\times n}$ be positive definite. It is trivial for $n=1$, just take the square root. Assume that a Cholesky factorization exists for positive definite matrices of dimension $n-1$ and partition $A$ as
$$
A
=
\begin{bmatrix}
\tilde{A}&a\\a^T&\alpha
\end{bmatrix},
$$
where $\tilde{A}\in\mathbb{R}^{(n-1)\times(n-1)}$. Since a principal submatrix of a positive definite matrix is positive definite, $\tilde{A}$ has a Cholesky factorization $\tilde{A}=\tilde{L}\tilde{L}^T$. Consider
$$\tag{1}
L_1^{-1}AL_1^{-T}
:=
\begin{bmatrix}
\tilde{L}^{-1}&0\\
0&1
\end{bmatrix}
\begin{bmatrix}
\tilde{A}&a\\
a^T&\alpha
\end{bmatrix}
\begin{bmatrix}
\tilde{L}^{-T}&0\\
0&1
\end{bmatrix}
=
\begin{bmatrix}
I&b\\
b^T&\alpha
\end{bmatrix}
=:B,
\quad b:=\tilde{L}^{-1}a.
$$
Next we eliminate $b$ by
$$\tag{2}
L_2^{-1}BL_2^{-T}
:=
\begin{bmatrix}
I&0\\-b^T&1
\end{bmatrix}
\begin{bmatrix}
I&b\\
b^T&\alpha
\end{bmatrix}
\begin{bmatrix}
I&-b\\0&1
\end{bmatrix}
=
\begin{bmatrix}
I&0\\0&\alpha-b^Tb
\end{bmatrix}
=
\begin{bmatrix}
I&0\\0&\alpha-a^TA^{-1}a
\end{bmatrix}.
$$
The diagonal matrix on the right-hand side of (2) is a result of congruence transformations applied to $A$, so the right-hand side of (2) is positive definite and $0<\alpha-a^TA^{-1}a=\lambda^2$ for some real $\lambda$. Set
$$
L_3:=\begin{bmatrix}I&0\\0&\lambda\end{bmatrix}
$$
so $L_2^{-1}BL_2^{-T}=L_3L_3^T$. From (1) we have
$$
L_2^{-1}L_1^{-1}AL_1^{-T}L_2^{-T}=L_3L_3^T,
$$
so
$$
A=LL^T,
\quad
L:=L_1L_2L_3
=
\begin{bmatrix}
\tilde{L}&0\\
0&1
\end{bmatrix}
\begin{bmatrix}
I&0\\b^T&1
\end{bmatrix}
\begin{bmatrix}
I&0\\
0&\lambda
\end{bmatrix}
=
\begin{bmatrix}
\tilde{L}&0\\
b^T&\lambda
\end{bmatrix}
=
\begin{bmatrix}
\tilde{L}&0\\
a^T\tilde{L}^{-T}&\lambda
\end{bmatrix}
$$
is a Cholesky factorization of $A$.
There is a source of non-uniqueness of the factorization in the choice of the sign of $\lambda$. As soon as one requires the signs of the diagonal terms of the Cholesky factors to be fixed (e.g., positive), the factorization is unique.
A simple way to confirm this can be made as follows. Assume
$$
A=LL^T=MM^T
$$
are two Cholesky factors of $A$. This gives
$$\tag{3}
I=L^{-1}MM^TL^{-T}=(L^{-1}M)(L^{-1}M)^T
$$
and
$$\tag{4}
(L^{-1}M)=(L^{-1}M)^{-T}.
$$
The left-hand and right-hand sides of (4) are, respectively, lower and upper triangular matrices which means that $D:=L^{-1}M$ is both lower and upper triangular and hence a diagonal matrix. From (3) we have $I=D^2$ so $D$ is a diagonal matrix with $\pm 1$ diagonal entries and $M=LD$ meaning that two Cholesky factors of $A$ differ by the signs of their columns.
