Prove that $0!+1! + 2! + 3! + ..... + n!$ $\neq$ $p^\text{r}$, where $n \geqslant 3$ and $n$, $p$ and $r$ are three integers

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.

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.