Prove that $0!+1! + 2! + 3! + ..... + n!$ $\neq$ $p^\text{r}$, where $n \geqslant 3$ and $n$, $p$ and $r$ are three integers 
Let $n$, $p$ and  $r$ be three positive integers. Prove that for $n \geqslant 3, r>1$, $$\sum_{k = 0}^{n} k! \neq p^\text{r}$$

SOURCE: BANGLADESH MATH OLYMPIAD (Preaparatory Question)
I am not so familiar with such a this kind of problem. Seeing that problem, I became little bit curious about what the text states and what will be its conception? 
I couldn't solve the problem. Moreover, I don't know about the formula of $\sum k!$. Is there any? And I couldn't realize the essence of $p^\text{r}$ and what the reason is behind the fact the summation can't be equal to $p^r$ for some integer $p$.
Any kind of reference or conception will massively help me start with some approach to solve the above problem. Thanks for your support and effort in advance.
 A: Just for fun, I will solve the same problem but with $0!$ removed:

Solve: 
  $$1!+2!+\dots+n!=p^r$$
  ...for $n,p,r\in N, r\ge2, n\ge3$

Denote the sum of the first $n$ factorials with $a_n$:
$$a_n=\sum_{k=1}^n k!$$
It is obvious that for $n\ge3$:
$$3\mid a_n$$
Why is that so? 3 divides 1!+2! and 3 divides all factorials starting from 3! So the sum of factorials $a_n$ must be divisible by 3 for $n>3$. It means that the $p^r$ must be divisible by 3 which is possible only if:
$$3\mid p\iff p=3q$$
In other words:
$$a_n=\sum_{k=1}^n k!=3^rq^r\tag{1}$$
Suppose that $r>=3$. It means that the right hand side is divisible by 27 and therefore it means that $a_n$ has to be divisible by 27 as well. Notice that 27 divides all factorials starting from $9!$. So if $a_8$ is divisible by 27, so it is $a_9, a_{10},\dots$. 
You can check manually that $$a_8=1!+2!+3!+4!+5!+6!+7!+8!=46233$$
Therefore:
$$27\nmid a_8$$
Consequentially, for all $n\ge9$:
$$27\nmid a_n$$
So we have a contradiction for all big enough values of $n$. We just have to check a few starting values manually. Indeed, if you check all values from $a_1$ to $a_8$ you will see that only $a_7=5913$ is divisible by 27. But number $5913=3^4\times73$ is not of the form $p^r$. 
So there is no solution for any $r\ge3$. 
We have yet to consider a case for $r=2$:
$$a_n=p^2$$
It can be easily shown (see proof here) that:
$$1!+2!+3!=3^2$$ 
...is the only solution of the problem.
A: The question asks about the prime factorisation of $\sum_{k=0}^{n}k!$. The first thing to notice is that you have $0!=1!=1$ and all the other terms are divisible by 2, so the sum is divisible by 2. Thus if we are looking for a counterexample, we must have $p$ even, and so since $r>1$, $4$ divides the sum.
Now, the sum begins $0!+1!+2!+3!=10$, and then all terms after this are $k!$ for $k\ge4$. Thus, for $n\ge3$, $4$ does not divide the sum, a contradiction.
