A bipartite pure state $\rho=|\psi\rangle\langle\psi|\in End(H_A\otimes H_B)$ was not entanglement if there were $|\psi_A\rangle\in H_A$ and $|\psi_B\rangle\in H_B$ such that $|\psi\rangle=|\psi_A\rangle\otimes|\psi_B\rangle$.
A general bipartite state $\rho\in End(H_A\otimes H_B)$ is separable if we can write it as a convex combinaion of product states such as $\rho=\sum_{i=1}^kp_i\sigma_A^{(i)}\tau_B^{(i)}$ where $\sigma_A^{(i)}\in End(H_A)$ and $\tau_B^{(i)}\in End(H_B)$. By convex combination I mean $\sum_{i=1}^kp_i=1$, $\forall 1\leq i\leq k\;:\;0\leq p_i\leq 1$. If $\rho$ is not separable we call it entanglement.
Now, I want to see if these two definitions are equal in case $\rho$ is pure.
So we take a pure state. One side is clear. Assume $\rho$ is not entanglement with the first definition then $$\exists\;|\psi_A\rangle\in H_A,|\psi_B\rangle\in H_B\;s.t.\;|\psi\rangle=|\psi_A\rangle\otimes|\psi_B\rangle$$ so $$\rho=|\psi_A\rangle|\psi_B\rangle\langle\psi_A|\langle\psi_B|=|\psi_A\rangle\langle\psi_A|\otimes|\psi_B\rangle\langle\psi_B|$$ Taking $k=1$, $p_1=1$, $\sigma_A^{(1)}=|\psi_A\rangle\langle\psi_A|$ and $\tau_B^{(1)}=|\psi_B\rangle\langle\psi_B|$, $\rho$ is seperable and not entanglement with the second definition.
But how can I show the inverse?