It is not that many!

Testing for small values of n is a good approach here, because only the factors included in the result can still be a solution to $k$ .

For $n=2$ : $2^7 - 2 = 128 - 2 = 126 = 2*3^2*7$ .

Now we only have to test, if the prime factors $2$ , $3$ and $7$ are always contained in $n^7 - n$ , and if the prime factor $3$ can occur twice.

Testing for divisibility by 2:

Case $n\equiv 0\ (\textrm{mod}\ 2)$ : $0^7 - 0 \equiv 0\ (\textrm{mod}\ 2)$

Case $n\equiv 1\ (\textrm{mod}\ 2)$ : $1^7 - 1 \equiv 0\ (\textrm{mod}\ 2)$

Therefore, the factor $2$ is always contained in $n^7 - n$ .

Similarily, you can test for the prime factors $3$ and $7$ . Then you can see that these factors are also always contained in $n^7 - n$ . Factoring $n^7 - n$ to $n(n^3-1)(n^3+1)$ helps seeing those factors.

Now we only have to check, whether the prime factor $3$ is always contained twice.

For the case $n\equiv1\ (\textrm{mod}\ 3)$ , according to the factorization $n(n^3-1)(n^3+1)$ of $n^7 - n$ , the factor $3$ is only contained once in $n^7 - n$ , namely in $(n^3-1)$ . Therefore the only prime factors that occur for all $n$ are $2$ , $3$ and $7$ .

The amount of numbers that can be made with the prime factors $2$ , $3$ and $7$ is eight, because every prime factor can either be part of the final number or not. Since every prime factor can be toggled, and there are three possible prime factors, the amount of numbers that can result from these prime factors alone is $2^3 = 8$ .

We can break $n^7 - n$ into polynomial factors as shown below

$n^7 - n = n (n^6 - 1) = n (n^3 -1)(n^3 + 1) = n (n -1) (n^2+n+1) (n+1) (n^2 - n + 1)$

Note that the above expression contains factors $n , (n+1) ,(n-1)$ which are three consecutive non-negative integers for every positive integer $n$ . It may be directly concluded that at least one of these integers has to be divisible by $3$ and also that atmost two of them will definitely have the opposite parity making one of the integers even and thus divisible by $2$ . Thus, we find that $n^7 - n$ must be divisible by $6$ .

Also, by Fermat's Little Theorem , we conclude that $7 | n^7 - n$ for every integer $n$ as $7$ is a prime number.

Therefore, $n^7 - n$ is divisible by $42$ and all of its positive factors which are: $\{1,2,3,6,7,14,21,42\}$ , giving us a total of $\boxed{8}$ values for $k$ .

By expanding, we get $n^7 -n = (n-1)(n)(n+1)(n^2 + n + 1)(n^2 - n +1)$ . We can clearly see that $n^7-n$ contain a product of three consecutive integer. Hence, it is divisible by $6$ .

Now, let $f(n) = n^7 - n$ . For $n= 2,$ we have $f(2) = 126$ which is divisible by $7$ . Now, we want to prove that $f(n+1) = (n+1)^7 - (n+1)$ is divisible by 7.

$\begin{aligned} (n+1)^7 - (n+1) & = \left( \displaystyle \sum_{i=0}^7 \space _7C_i \cdot n^{7-i} \right) - (n+1) \\ & = \color{#D61F06}n^7 \color{#333333} + \left( \displaystyle \sum_{i=1}^5 \space _7C_i \cdot n^{7-i} \right) + \left( \color{#3D99F6} \displaystyle \sum_{i=6}^7 \space _7C_i \cdot n^{7-i} \color{#333333} \right) \color{#D61F06} - n \color{#333333} - 1 \\ & = \left( \displaystyle \sum_{i=1}^5 \space _7C_i \cdot n^{7-i} \right) + \color{#D61F06} (n^7 - n) \color{#333333} + \color{#3D99F6} 7n + 1 \color{#333333} - 1 \\ & = \underbrace{ \left( \displaystyle \sum_{i=1}^5 \space _7C_i \cdot n^{7-i} \right)}_{ \text{Divisible by 7, because it contains } \space _7C_i} + \underbrace{ \color{#D61F06} (n^7 - n)}_{\color{#333333} \text{Always divisible by 7} } + 7n \end{aligned}$

We get, $7|n^7 - n$ .

Note that, if $6| n^7 - n$ and $7 |n^7 - n$ , then $42|n^7 - n$ .

Since $42$ has $8$ divisors, then the answer is $8$ , for $k = (1,2,3,6,7,14,21,42)$ .