$n!+k$ is never any power of any prime number if $n\ge 6$ and $2\le k\le n$? Question : Is the following true?
"If $n!+k$ is a power of a prime number, then it is one of $2!+2, 3!+2, 3!+3, 4!+3, 5!+5$ where $n,k\in\mathbb Z$ satisfy $n\ge 2$ and $2\le k\le n$."
Motivation : The following is well known : 
1. A sequence $(n+1)!+k\ (k=2,3,\cdots,n+1)$ does not have any prime number for any $n\in\mathbb N$.
I've just got the following : 
2. A sequence $\{(n+1)!+(n+1)\}!+(n+1)!+k\ (k=2,3,\cdots,n+1)$ does not have any power of any prime number for any $n\in\mathbb N$.
After thinking about these sequences, I reached the above expectation. However, I can neither prove this expectation is true nor find any counterexample. Can you help?

Suppose that 
$$n!+k=p^r\ \ (r\ge 2).$$
Since it is obvious that $k$ is a power of $p$, let $k=p^s\ (1\le s\lt r).$ However, if $s\gt 1,$ then $n\ge k=p^s\gt p$ leads that $p^{s+1}$ is a divisor of $n!$. Noting that $s+1\le r$ and that $p^{s+1}$ is a divisor of $p^r$, this leads that $p^{s+1}$ is a divisor of $p^r-n!=p^s$, which is a contradiction. Hence, we get $s=1, k=p\le n$. By the same argument as above, we get $n\lt 2p$.
By the way, the number of a prime factor $2$ included in $n!\ (n\ge 2)$ is $\sum_{i=1}^{\infty}\lfloor{\frac{n}{2^i}}\rfloor$ where $\lfloor x\rfloor$ is the largest integer not greater than $x$, so we get
$$\begin{align}\sum_{i=1}^{\infty}\lfloor\frac{n}{2^i}\rfloor\ge\lfloor\frac n2\rfloor+\lfloor\frac n4\rfloor\ge\left(\frac n2-\frac 12\right)+\left(\frac n4-\frac 34\right)=\frac{3n-5}{4}\qquad(1)\end{align}$$
On the other hand, letting $r-1=2^mu$ where $m$ is non-negative integer and $u$ is odd, we get
$$p^{r-1}-1=(p^{2^m})^u-1=(p^{2^m}-1)\{(p^{2^m})^{u-1}+\cdots+(p^{2^m})^2+p^{2^m}+1\}.$$
Since $\{\ \}$ is odd, we get $e(m)=\sum_{i=1}^{\infty}\lfloor{\frac{n}{2^i}}\rfloor$ where $e(m)$ represents the number of a prime factor $2$ included in $p^{2^m}-1$. Since $p^{2^m}-1=(p^{2^{m-1}}+1)(p^{2^{m-1}}-1)$, we know that $e(m)=1+e(m-1)$. (This is because $p^{2^s}+1\not\equiv 0$ (mod $4$) for any $s\in\mathbb N$.) This leads
$$\begin{align}e(m)\le e(1)+m-1\ \ (m\ge 0)\qquad(2)\end{align}$$ 
Let's consider $e(1)$. Letting $p=2l+1\ (l\in\mathbb N)$, since $p^2-1=(p+1)(p-1)=(2l+2)(2l)$, we get $2^{e(1)}=\frac{2^2(l+1)l}{q}$. Since $l\le q$, we get $2^{e(1)}\le 2^2(l+1)=2(p+1)\le 2(n+1)$, which leads
$$\begin{align}e(1)\le \log_2(n+1)+1\qquad(3)\end{align}$$
Let's consider $m$. Since $u$ is odd, noting $r-1=2^mu,$ we get $2^m\le r-1\iff m\le \log_2(r-1).$ Since $p^r\lt p+2p^n\le 3p^n\le p^{n+1}$, we get $r\le n$. Here I used $n!\le 2\left(\frac n2\right)^n\lt 2p^n$ since $n\lt 2p$. Hence we get 
$$\begin{align}m\le \log_2(n-1)\qquad(4)\end{align}$$
From $(1)(2)(3)(4)$, we get
$$\frac{3n-5}{4}\le \log_2(n^2-1).$$
To make calculations easier, let's consider $\frac{3n-5}{4}\lt \log_2n^2$. Then, we know that $n$ must satisfy the following condition :
$$3n-5\lt 8\log_2n,\ n\ge 3.$$
Observing the function $f(x)=8\log_2x-3x+5=8\frac{\log_ex}{\log_e2}-3x+5\ (x\ge 3),$ we get $3\le n\le 10.$ Observing every $n$, we know that $2!+2, 2!+2, 3!+3,4!+3,5!+5$ are the only such examples. Now the proof is completed.
