If $A$ is symmetric, then the matrix exponential $e^{A}$ is positive definite

If $A$ is symmetric, it is diagonalizable. So write $$A = PDP^{-1}$$ where $D = \text{diag}(\lambda_1, ..., \lambda_n)$ is a diagonal matrix whose entries are the eigenvalues of $A$. Thus $$e^A = Pe^DP^{-1}$$ and it is easy to see that $e^D = \text{diag}(e^{\lambda_1}, ..., e^{\lambda_n})$. But since the exponential is always positive, this means that all the eigenvalues of $e^A$ are positive. Hence $e^A$ is positive definite.

Since $A$ is symmetric and the operator $\cdot^T\colon A\in \mathcal M_n(\mathbb R)\mapsto A^T\in\mathcal M_n(\mathbb R)$ is continuous (as a linear operator in a finite dimensional vector space), the matrix $e^{A/2}$ is symmetric. Therefore, we have for $x\in\mathbb R^n$: $$x^Te^Ax=x^Te^{A/2}e^{A/2}x=x^T(e^{A/2})^Te^{A/2}x =(e^{A/2}x)^Te^{A/2}x=\lVert e^{A/2}x\rVert^2\geq 0,$$ and since $e^{A/2}$ is invertible, we have the equality if and only if $x=0$, which shows that $e^A$ is positive definite.

I assume that $A$ is a real matrix here. Here is a way that requires some knowledge about linear algebra but no calculation. By spectral theorem $A=U \Lambda U^{-1}$ where $\Lambda$ is the diagonal matrix, the entries of which are the eigenvalues of $A$ (hence non negative), call these $\lambda_i\geq 0$. Take now $e^A$ which is clearly symmetric, its eigenvalues are $e^{\lambda_i} > 0$, and you are done.