Experiment:-$\sum_{j=1}^{2n+1}(j+1)\left\lfloor {j\cdot j!\over j+1} \right\rfloor$ Experiment of Mathematica to $(1)$

$$\sum_{j=1}^{2n+1}(j+1)\left\lfloor {j\cdot j!\over j+1} \right\rfloor \tag1$$
  $\pi(n)$;Prime-counting function
$\lfloor x \rfloor$;Floor function

We spotted a closed form given in term of Prime-counting function $\pi(n)$
We would like to know if it is correct for all values of $n\ge1$

How can we show that $(1)=(2n+2)!-\pi(2n+1)-3$?

Greatly appreciated, if you can help.
Example
$n=1$

$$\sum_{j=1}^{3}(j+1)\left\lfloor {j\cdot j!\over j+1} \right\rfloor=2\lfloor (1/2)\rfloor+3\lfloor (4/3)\rfloor+4\lfloor (18/4)\rfloor=0+3+16=19$$

$19=4!-\pi(3)-3=24-2-3$
where $\pi(3)=2$
 A: We will use the fact that $\lfloor n+a\rfloor=n+\lfloor a\rfloor$ if $n\in \mathbb{N}$.
We can see that $j\cdot j!=(j+1)!-j!$.Dividing with $j+1$ yields $\frac{j\cdot j!}{j+1}=\frac{(j+1)!}{j+1}-\frac{j!}{j+1}$.
 So, the formula takes the form $$\sum_{j=1}^{2n+1}(j+1)\left\lfloor {j\cdot j!\over j+1} \right\rfloor=\sum_{j=1}^{2n+1}(j+1)\left\lfloor\frac{(j+1)!}{j+1}-\frac{j!}{j+1} \right\rfloor,$$ which is equal to $$\sum_{j=1}^{2n+1}(j+1)!-(j+1)\left\lfloor {j!\over j+1} \right\rfloor.$$
If $j+1$ is composite $>4$ then $j+1$ divides $j!$ so $\left\lfloor {j!\over j+1} \right\rfloor=\frac{j!}{j+1}$ which means that $(j+1)\left\lfloor {j!\over j+1} \right\rfloor=j!$.
On the other hand, if $j+1$ is prime then, from Wilson's theorem we know that $j!\equiv -1\pmod{j+1}$ which gives $\left\lfloor {j!\over j+1} \right\rfloor=\frac{j!+1}{j+1}$ and $(j+1)\left\lfloor {j!\over j+1} \right\rfloor=j!+1$.  
(This '$1$' left is very important since it occurs for all the $\pi(2n+1)$ primes)    
Combining all from above, you can rewrite your sum as $$\sum_{j=1}^{2n+1}(j+1)!-j!-\pi(2n+1)-2.$$ Note that the '$2$' comes because of the special composite $4$ we mentioned.
This sum is a telescoping one and it is equal to $$(2n+2)!-\pi(2n+1)-3.$$
