Statement-1: ${(n+1)}^7-n^7-1$  is divisible by 7,  $\forall n\in N$  .
Statement-2: $n^7-n$  is divisible by 7,  $\forall n\in N$ .

Consider statement 2.
$p\left(n\right)=n^7-n$   Is divisible by 7. $\forall n\in N$ 
$\therefore p\left(1\right)=1^7-1=0$  is divisible by 7.
$p\left(1\right)=1$  is true
Suppose $p\left(k\right):k^7-k$  is divisible by 7 true. Let  $k^7-k=7m,m\in N\left(1\right)$
 $$\begin{gathered} {(k+1)}^7-(k+1)=k^7+\left(\begin{array}{l}7\\ 1\end{array}\right)k^6+\left(\begin{array}{l}7\\ 2\end{array}\right)k^5+\mathrm{....}+7k+1-k-1 \\ =(k^7-k)+\left(\begin{array}{l}7\\ 1\end{array}\right)k^6+\left(\begin{array}{l}7\\ 2\end{array}\right)k^5+\mathrm{.....}+\left(\begin{array}{l}7\\ 6\end{array}\right)k \end{gathered}$$
$k^7-k$  is divisible by 7 and $\left(\begin{array}{l}7\\ r\end{array}\right)$  is divisible by 7. For  $1\le r\le 6. \left(By\left(1\right)\right)$
$\therefore p(k+1)$  is true.
$\therefore p\left(n\right)$  is true, $\forall n\in N$ .
$\therefore$  Statement (2) is true.
Now consider statement 1.
From statement 2,  $n^2-n$   is divisible by 7.
$\therefore {(n+1)}^7-(n+1)$   is divisible by 7.
$\therefore {(n+1)}^7-(n+1)=7m,m\in N$  and   $n^7-n=7p. p\in N$ 
 $$\begin{gathered} \therefore {(n+1)}^7-n^7+n^7-n-1=7m \\ \therefore [{(n+1)}^7-n^7-1]+(n^7-n)=7m \\ \therefore {(n+1)}^7-n^7-1=7(m-p) \end{gathered}$$
$\therefore {(n+1)}^7-n^7-1$   is divisible by 7.       $\forall n\in N$
$\therefore$  Statement1 is also true.
$\therefore$   Statement 1 and statement 2 both are true and statement 2 is the correct explanation for statement 1.