Swap, delete and repeat Introduction
Let's observe the following string:
ABCDEFGHIJKLMNOP

If we swap the ends of the string, which are these:
ABCDEFGHIJKLMNOP
^^            ^^

We get the following result:
BACDEFGHIJKLMNPO

After that, we delete the ends of the string, which in this case are B and O. The result is:
ACDEFGHIJKLMNP

If we repeat the process, we get the following list:
N     Result

2     ADEFGHIJKLMP
3     AEFGHIJKLP
4     AFGHIJKP
5     AGHIJP
6     AHIP
7     AP

You can see that for N = 5, the result is AGHIJP. At N = 7, the length of the string is smaller than 3, so N &gt; 7 is considered invalid in this case.
The Task
Given a string S with at least length 4 and the number of repetitions N &gt; 0, output the final result. You can assume that N is always valid.
Test cases
Input                               &gt;  Output

N = 3, S = ABCDEFGHIJKLMNOP         &gt;  AEFGHIJKLP
N = 1, S = Hello                    &gt;  Hlo
N = 2, S = 123321                   &gt;  11

This is code-golf, so the submission with the least amount of bytes wins! For the simplicity, you can assume that the string will only contain alphanumeric characters.
Retina, 44 20 bytes

Crossed out 44 is still regular 44 :(

+`'(\w).(.*).\B
$1$2


Assumes input in following format (in unary - counting character: '):
{number of repeats}{string}
For example: '''''''ABCDEFGHIJKLMNOP
There is no space between number of repeats and the string.

Thanks @MartinBüttner for shaving off 24 bytes!

Try it online!