There is 1 power of $n$ between $(n)!$ and $(n+1)!$ except for $n=5$ ? same for primorial? Is the following assertion true? 

There is 1 power of $n$ between $(n)!$ and $(n+1)!$ except for $n=5$ ?

For $n=5$, there are 2 powers of $5$ between $5!$ and $6!$ which are $5^3$ and $5^4$
For other cases we have ($k={\lfloor \log_nn!\rfloor}$):
$$n^{k-1}<(n-1)!<n^k<(n)!<n^{k+1}<(n+1)!<n^{k+2}$$
The left part is easy to show, but not the right end (I guess)
Now, another one
Is the following assertion true? 

There is 1 power of $p_i$ between $(p_i)\#$ and $(p_{i+1})\#$ except for $p_i$ in $\{3,557,46187\}$ ?

e.g. for $p_i=557$, there are 2 powers of $557$ between $557\#$ and $563\#$ which are $557^{82}$ and $557^{83}$
For other cases we have ($k={\lfloor \log_{p_i}(p_i)\#\rfloor}$):
$$p_i^{k-1}<(p_{i-1})\#<p_i^k<(p_i)\#<p_i^{k+1}<(p_{i+1})\#<p_i^{k+2}$$
The left part is easy to show, but not the right end (I guess)

Is there a way to show the first or the second. Can we use the same logic for the other?

 A: The event that there are $2$ powers of $n$ between $n!$ and $(n+1)!$ is the same as saying that
$$
\bigg\lfloor \frac{\log((n+1)!)}{\log n}\bigg\rfloor - \bigg\lfloor \frac{\log(n!)}{\log n}\bigg\rfloor = 2.
$$
Since the raw difference is
$$
\frac{\log((n+1)!)}{\log n} - \frac{\log(n!)}{\log n} = \frac{\log(n+1)}{\log n} \approx 1 + \frac1{n\log n},
$$
this is only possible (as alex.jordan pointed out) if the fractional part of $\frac{\log(n!)}{\log n}$ is extremely close to $1$; indeed it should be greater than $\approx 1 - \frac1{n\log n}$.
If we treat this as a random event with probability $\approx \frac1{n\log n}$, then as we examine all $n$ we could encounter the series $\sum_{n=2}^\infty \frac1{n\log n}$, which diverges by the integral test. By analogy with the Borel–Cantelli lemma, we would actually predict that there are infinitely many $n$ for which there are $2$ powers of $n$ between $n!$ and $(n+1)!$. (The number of such $n$ would grow extremely slowly—the $k$th such one might be around $e^{e^k}$ by this heuristic.)
A similar analysis holds for primes: we need the fractional part of $\frac{\log(p_j\#)}{\log p_j}$ to be larger than $\approx 1 - \frac{p_{j+1}-p_j}{p_j\log p_j}$, which has "probability" $\approx \frac{p_{j+1}-p_j}{p_j\log p_j}$; and this series also diverges (essentially because the average value of $\frac{p_{j+1}-p_j}{\log p_j}$ equals $1$ by the prime number theorem).
