Permutations in an k-sorted array Definition of $k$-sorted array: An array in which an element is at-most $k$ places away from its sorted order.

I have a question in my Algorithms assignment which asks to prove the lower bound to sort a $k$-sorted array as $\Omega(n\log{k})$. I was trying to approach this question by using the standard comparison sort proof ($\Omega(n!)$), where we have $n!$ total permutations of the sequence [$a_1,a_2,\cdots,a_n$]. My friend told me that there are $k^{(n-k)}k!$ permutations of a $k$-sorted array. I'm finding it difficult to prove the same. How should I go about finding the total permutations of a $k$-sorted array?

Try for $n=5$, $k=2$.

Total permutation is $3\times 3\times 3\times 2\times 1$, 
First element has 3 possibilities $\{1,2,3\}$; second has 4 possibilities $\{1,2,3,4\}$, but one position is already taken by the first,so effectively it has $(4-1)$ slots; third has 5 possibilities $\{1,2,3,4,5\}$, but first and second are there in one of these two, hence 3 possibilities; Fouth one has 4 slots $\{2,3,4,5\}$ and 3 of them are taken? (doubt: because 1st position can also be occupied) , hence 2 , and the last has one.
Let $N(n,k)$ be the number of $k$-sorted arrays on $n$ elements. At each position we have at most $(2k+1)$ possible elements, giving an upper bound of $(2k+1)^n$. On the other hand, suppose that we divide $\{1,\ldots,n\}$ into $n/k$ blocks of size $k$ (assuming for simplicity that $k$ divides $n$), and apply an arbitrary permutation on each block. All of these arrays are $k$-sorted, showing that $N(n,k) \geq k!^{n/k}$. In total,
$$
k!^{n/k} \leq N(n,k) \leq (2k+1)^n \\
\frac{n}{k} (k \log k - k + O(\log k)) \leq \log N(n,k) \leq n \log(2k+1)
$$
We conclude that $\log N(n,k) = \Theta(n \log k)$.

For the exact expression, consult a paper of Kløve or the OEIS.