You can prove it easily with the (already mentioned) Cauchy interlacing theorem, the fact that the determinant is equal to the product of eigenvalues and that the (real) symmetric matrix is positive definite if and only if its eigenvalues are positive.
The statement is obviously true for $1\times 1$ matrices. Assume that "all leading principal minors of $A$ are positive implies $A$ is positive definite" is true for $k\times k$ matrices, $k\leq n-1$, and consider $A$ to be $n\times n$ with all leading principal minors positive. By the induction assumption, we know that the leading principal $(n-1)\times(n-1)$ submatrix of $A$ is positive definite. By the interlacing property, all "larger" $n-1$ eigenvalues of $A$ are positive up to (possibly) the smallest one. However, the smallest eigenvalue cannot be nonpositive since otherwise the determinant of $A$ would not be positive.