For any positive integer $t>1$, we can define $\nu_t(n)$ as the maximal integer $k$ such that $t^k\mid n$. (Here we don't need $t$ to be a prime.)
Then I wonder if we have $$\lim_{t\rightarrow\infty}\frac{\nu_t(t!)}{t}=0$$
For the most naive bound, I only have $$\nu_t(t!)\leq \log_t (t!)\sim t$$ So it must use Legendre's theorem to delete some useless prime in $t!$. But I still cannot prove this limit...
$$ \frac{\nu_t(t!)}{t} \leqslant \frac{\nu_p(t!)}{\nu_p(t) \cdot t} \leqslant \frac{t}{(p-1)\nu_p(t)\cdot t} = \frac{1}{(p-1)\nu_p(t)} $$
for any prime $p$ (where first inequality follows from the fact that $t^{\nu_t(t!)} \mid t!$, and the second one is consequence of Legendre's formula. Suppose $\frac{\nu_t(t!)}{t}$ will stay above some $c>0$ infinitely many times, so our sequence won't be convergent to zero. Then, for all $p \mid t$, $c \leqslant \frac{1}{p-1}$, so $p \leqslant c^{-1}+1$ and all prime divisors of $t$ are bounded. Similarly, for all $p \mid t$ we have $\nu_p(t) \leqslant c^{-1}$. But this gives bound on $t$: it can't be larger than product of all primes less than $c^{-1}+1$, raised to power $c^{-1}$. This is a contradiction with assumption about our sequence not being convergent to zero; thus
$$ \lim_{t \to \infty} \frac{\nu_t(t!)}{t} = 0 $$
as we wanted to prove.