Every neighborhood of identity in a topological group contains the product of a symmetric neighborhood of identity.

Every neighborhood of identity in a topological group contains the product
of a symmetric neighborhood of identity.

Let $(G,\cdot)$ be a topological group and $U$ be a neighborhood of $1$.
Then there exists a symmetric neighborhood of $1$, $V^{-1} = V$, such that
$V\cdot V \subset U$. Having a hard time proving this. $V^{-1} = \{v^{-1}:
v \in V\}$. I know that an open set times any set is also open. And a hint
is that $VV^{-1}$ is symmetric and an open neighborhood of $1$ when $V$ is
an open neighborhood of $1$ contained in $U$, but showing $VV^{-1}$ or an
expression involving it is a subset of $U$ or $V$ requires something else.