The Squaring Sequence

JavaScript (ES7), 44 43 36 bytes

f=n=>--n?(f(n)**2+f).slice(0,4):1111

This is a great example of abusing type coercion: ** converts both its arguments to numbers, and + converts both its arguments to strings unless they're both numbers. This means that f(n)**2+f first converts f(n) to a number and squares it, then concatenates the result with the string representation of f. We can then use .slice to retrieve the first 4 chars of the string.

Here are a few alternate approaches that don't use strings:

f=(n,x=1111)=>x<1e4?--n?f(n,x*x):x:f(n,x/10|0)
f=n=>--n?(x=f(n))*x/(x>3162?1e4:1e3)|0:1111

Test snippet

let f=n=>--n?(Math.pow(f(n),2)+f).slice(0,4):1111

<input id=I type="number" step="1" min="1" value="1"><button onclick="console.log(f(I.value))">Run</button>

Note: this uses Math.pow because ** isn't supported in all browsers.

05AB1E, 8 7 bytes

Code:

$Fn4×4£

Explanation:

$        # Push 1 and the input
 F       # Input times do...
  n      #   Square the number
   4×    #   Repeat that string 4 times
     4£  #   Take the first four characters
         # Output the last computed number

Uses the CP-1252 encoding. Try it online!

Python 2, 51 46 44 Bytes

I'd like to get rid of the clunky if if possible, but I think an exec could be shorter.. Turns out for the moment that exec is shorter. Wrong again! The recursive function returns. This is one-indexed.

f=lambda n:1111*(n<2)or int(`f(n-1)**2`[:4])

An aleternative 46-byte solution with exec:

s=1111;exec's=int(`s*s`[:4]);'*input();print s

An alternative 49-byte recursive solution:

f=lambda n,s=1111:s*0**n or f(n-1,int(`s*2`[:4]))

^{Thanks to Flp.Tkc for saving a byte by reminding me that squaring doesn't need exponentiation :)}