Is there a prime $p$ and an infinite sequence $a_1<a_2<\cdots$ such that if $p^{m_k}$ is the highest power of $p$ dividing $\sum_{i\leq k}a_i!$ then $m_k\to \infty$?

OPEN

Let $f(a,p)$ be the largest $k$ such that there are $a=a_1<\cdots<a_k$ such that
\[p^k \mid (a_1!+\cdots+a_k!).\]
Is $f(a,p)$ bounded by some absolute constant? What if this constant is allowed to depend on $a$ and $p$?

Is there a prime $p$ and an infinite sequence $a_1<a_2<\cdots$ such that if $p^{m_k}$ is the highest power of $p$ dividing $\sum_{i\leq k}a_i!$ then $m_k\to \infty$?