Assuming n is an integer, there are no values of n so that 2^n is a multiple of 10 because 2^n doesn't contain any necessary factors of 5.
The unit digits of 2^n for n=1,2,3... are 2, 4, 8, 6 then repeat. For 3^n they go 3, 9, 7, 1 then repeat. If n is odd, the units that are being added are either 2 + 3 or 7 + 8, which both end in 5. So 2^n + 3^n where n is odd always ends in 5. This is a stronger conclusion than saying 'it's a multiple of 5' as a multiple of 5 can also end in 0.
If n is a multiple of 4 the units being added are 6 + 1 so in this case it will always end in 7.
1^n + 2^n + 3^n is even for all values of n. This is obvious because 1^n and 3^n are always odd and 2^n is always even, and odd + odd + even = even.
1^n + 2^n + 3^n + 4^n is a multiple of 10 for 1, 2 or 3 mod 4.
The unit digits of the powers of 1, 2, 3 and 4 are:
1 1 1 1
2 4 8 6
3 9 7 1
4 6 4 6
Summing down the columns gives 10, 20, 20 and 14. This shows that when n is 1,2 or 3 mod4, the last digit of 1^n + 2^n + 3^n + 4^n is 0 so it's a multiple of 10.
1^n + 2^n + 3^n + 4^n + 5^n ends in 5 for 1, 2 or 3 mod4. This is obvious if we consider the previous result because 5^n ends in 5 for all n, so adding this to the multiples of 10 will give a multiple of 5.