Determinant of a matrix with binomial coefficients. Let $n \in\mathbb{N}$ and $A=(a_{ij})$ where 
\begin{equation}a_{ij}=\binom{i+j}{i}\end{equation}
for $0\leq i,j \leq n$. Show that $A$ has an inverse and that every element of $A^{-1}$ is an integer.
I have shown that this $n\times n$ matrix is symmetric since, 
\begin{equation}
\binom{i+j}{i}=\binom{i+j}{j}
\end{equation} 
in order to try to get a nonzero determinant. 
But i'm stuck in this step, suggestions would be appreciated.    

If you take a particular case (say $n=5$) and you consider the LDL or Cholesky decomposition of this matrix you notice something very interesting: ( WA link).
So one should try to prove that our matrix $A$ is the product
$$A = L \cdot L^t$$
where
$L = (\binom{i}{j})_{0\le i,j \le n}$.This translates into the equalities:
$$\sum_k\binom{i}{k}  \binom{j}{k} = \sum_k\binom{i}{k}  \binom{j}{j-k}= \binom{i+j}{j}$$
Note that $L$ is lower triangular with $1$ on the diagonal, so invertible. In fact, its inverse can be explicitely given
$$L^{-1} = ((-1)^{i-k} \binom{i}{j})$$
One can prove a more general equality
$$L^{a}\cdot L^{b}=L^{a+b}$$
where $L_{ij}^a =a^{i-j} \binom{i}{j}$.
So now we get
$$A^{-1} = (L\cdot L^t)^{-1}= (L^{-1})^t \cdot L^{-1}$$
so clearly with integral elements, although I can't notice a particularly nice form for the entries.
Notice: matrices of the form $\binom{a+i+j}{j}$, or $\binom{a+i}{j}$, or $\binom{a-i}{j}$, or more generally, $\binom{a \pm i + b j}{j}$ also have nice LU decompositions. ($a$, $b$  parameters). The L part (the lower triangular part) is always $\binom{i}{j}$.
