Power-Ending Patterns When you look at the last decimal digit of each power of a non-negative integer a repeating pattern forms. For example, 3:
3^1 =     3
3^2 =     9
3^3 =    27
3^4 =    81
3^5 =   243
3^6 =   729
3^7 =  2187
3^8 =  6561
3^9 = 19683

The last digits go 3971 which repeats indefinitely. In fact any number we choose that ends in 3 has that same pattern because none of the other digits can have an effect on the ones place during repeated multiplication.
What's curious is that some numbers have a much shorter cycle of power-ending digits. For example with 5 the powers all end in 5 so the pattern, written as short as possible, is simply 5.
Looking at the minimal power-ending digits patterns for 0 through 9 we get:
0 -&gt; 0
1 -&gt; 1
2 -&gt; 2486
3 -&gt; 3971
4 -&gt; 46
5 -&gt; 5
6 -&gt; 6
7 -&gt; 7931
8 -&gt; 8426
9 -&gt; 91

(The lengths of these being 11442 repeated is a curious tidbit itself.)
Remember, any numbers above 9 will have the same pattern as their last digit as was explained above with 3.
Challenge
Your challenge here is to write a program that takes in any non-negative integer and outputs its minimal power-ending digit pattern.
The exact output formatting, whether string or list, doesn't matter. For example, here are some potential inputs followed by valid potential outputs:
900 -&gt; [0]
11 -&gt; 1
2 -&gt; 2486
303 -&gt; 3, 9, 7, 1
44 -&gt; 4 6
45 -&gt; 5
666 -&gt; &quot;6&quot;
3857 -&gt; [7 9 3 1]
118 -&gt; '8426'
129 -&gt; [9, 1]

The shortest code in bytes wins.
Python 3, 44 43 40 bytes

Based off of mabel's answer but using dictionary comprehensions:

lambda n:[*{n**i%10:0for i in[1,2,3,4]}]


Saved 1 byte thanks to mabel.

Saved 3 bytes thanks to xnor.

You can try it online