Colorful Power Sum

Looking at the last digits, $1^1 \equiv 1 \pmod{10} \\ 2^2 \equiv 4 \pmod{10} \\ 3^3\equiv 7 \pmod{10} \\ 4^4 \equiv 6 \pmod{10} \\ 5^5 \equiv 5\pmod{10} \\ 6^6\equiv 6\pmod{10} \\ 7^7\equiv 3\pmod{10} \\ 8^8 \equiv 6 \pmod{10} \\ 9^9\equiv 9\pmod{10} \\ 10^{10} \equiv 0 \pmod{10}$

Sum $\pmod{10}$ is $1+4+7+6+5+6+3+6+9 \equiv \boxed{7} \pmod{10}$

Looks like a good design made by $\LaTeX$ LOL

No calculator using only the last digit with maximum simplification,

1 ⇒ 1

2•2 ⇒ 4

3•3•3 ⇒ 7

4•4→6, 6•6 ⇒ 6

Multiplicating 5s will always finish by ⇒ 5

Multiplicating 6s will always finish by ⇒ 6

7•7→9,...•7→3, ...•7→1, period of 7931 ⇒ 3

8•8→4, 4•4→6, 6•6 ⇒ 6

9•9→1, odd period of 91 ⇒ 9

Multiplicating 10s will always finish by ⇒ 0

1+4+7+6+5+6+3+6+9+0, cancelling pairs 1+9, 4+6 and 7+3

6+5+6 = 17

Answer = 7

[It took me longer to write this than to solve the problem.]

Hooray for list comprehensions.

Python:

1

print str(sum([i**i for i in range(1,11)]))[-1]

$1^n$ will always be 1

Last digit of $2^{4n+1},2^{4n+2},2^{4n+3},2^{4(n+1)}$ is respectively $2,4,6,8$ . $2=4 \times 0+2$ so the last digit of $2^2$ is 4

Last digit of $3^{4n+1},3^{4n+2},3^{4n+3},3^{4(n+1)}$ is respectively $3,9,7,1$ . $3=4 \times 0+3$ so the last digit of $3^3$ is 7

Last digit of $4^{2n+1},4^{2(n+1)}$ is respectively $4,6$ . $4=2 \times 2$ so the last digit of $4^4$ is 6

Last digit of $5^n,6^n$ is respectively always $5,6$

Last digit of $7^{4n+1},7^{4n+2},7^{4n+3},7^{4(n+1)}$ is respectively $7,9,3,1$ . $7=4 \times 1+3$ so the last digit of $7^7$ is 3

Last digit of $8^{4n+1},8^{4n+2},8^{4n+3},8^{4(n+1)}$ is respectively $8,4,2,6$ . $8=2 \times 4$ so the last digit of $8^8$ is 6

Last digit of $9^{2n+1},9^{2(n+1)}$ is respectively $9,1$ . $9=2 \times 4+1$ so the last digit of $9^9$ is 9

Last digit of $10^n$ is always 0

So the sum of the last digits is $1+4+7+6+5+6+3+6+9+0=47$ .So the answer is $\boxed{\large{7}}$

This problem can be solved by just manually computing some smaller powers ( like till 4^4 ) and then using the concept of cyclicity ( which states that the unit's digit of n^k is the same as that of n^x where x is the residue of k modulo 4.)