Prove that $b_n = b_{n+100}$ 
Let $b_n$ denote the units digit of $\displaystyle\sum_{a=1}^n a^a$. Prove that $b_n = b_{n+100}$.

I tried rewriting the sum, but didn't see how to prove the equality. For example, if $n = 178$ we have $$\displaystyle\sum_{a=1}^{178} a^a = (1^1+2^2+3^3+\cdots+78^{78})+(79^{79}+80^{80}+81^{81}+\cdots+178^{178})$$ and so we need to show that $79^{79}+80^{80}+81^{81}+\cdots+178^{178} \equiv 0 \pmod{10}$. Is there an easier way to prove the general result?

It is sufficient to prove that for each integer $k$
$$\sum_{a=t+1}^{t+100} a^a \equiv 0 \mod10$$
Consider a subsequence with indices $k,k+10,\dots,k+90$. We have
$$
b(k) = k^k + (k+10)^{k+10} +\dots+(k+90)^{k+90}
$$
Obviously, the following holds $$b(k) \equiv k^k + k^{k+10} + \dots+k^{k+90}\mod10$$
The right side could be rewritten
$$k^k + k^{k+10} +\dots+k^{k+90} = k^k(1 + k^{10} +\dots+k^{90})$$

Lemma 1$$1 +k^{10} + k^{20}+\dots+k^{90} + k^{100} \equiv 1 \mod10$$ 
  Proof. Here we use Fermat's little theorem, which states that $b$ coprime with $10$ sutisfies the condition $b^\varphi(10) = b^4\equiv 1\mod 10$. And for $b$ coprime with $5$ we have $b^\varphi(5) = b^4\equiv 1\mod 10$.
  So, in the proof it is sufficient to check only the integers from $[0,9]$. 
  
  
*
  
*For the coprimes with $10$ we have $$1 +k^{10} + k^{20}+\dots+k^{100}\equiv 1 + 5(1+k^2)\equiv 1\mod10.$$ 
  
*For $k = 5$ we have $5^{10i}(1-5^{100-10i})\equiv 1\mod10$, where the expression inside the brackets is even. 
  
*For an even $k$ we get
  $$1 + k^{10}(1 + k^{10}+\dots+k^{90})\equiv 1 \mod10,$$
  where the expression in the brackets is divisible by $5$ because
  $$1 + k^{10}+\dots+k^{90}\equiv   5(1+k^2)\equiv 0\mod5,$$
  

Then 
$$k^k + k^{k+10} + k^{k+20}+\dots+k^{k+90} \equiv k^k(1-k^{100})\mod10$$

Lemma 2 $$a^a(1 - a^{100})\equiv 0\mod10,$$ for an integer $a$. 
Proof. 
  
  
*
  
*For the comprimes with $10$, it is straightforward from Fermat's little theorem. 
  
*For $a \equiv 5\mod10$ we have even value for $(1-a^{100})$. 
  
*For an even number $a$ comprime with $5$, the quantity $(1-a^{100})$ is divisible by $5$ according to Fermat's little theorem.
  
*For $a\equiv 0 \mod10$ it is obvious.

Finally, we can return to the initial sum and rewrite it
\begin{align}
\sum_{a=k}^{k+100}a^a &\equiv
k^k(1 - k^{100}) + (k+1)^{k+1}(1 - (k+1)^{100}) +\dots+(k+9)^{k+9}(1 - (k+9)^{100})\mod10
\end{align}
Where each term is divisible by $10$ according to Lemma 2.
Therefore, the whole last sum is actually divisible by $10$.
