How many zeroes are there at the end of the sum $1^1 + 2^2 + 3^3 + \cdots+ 100^{100}$?

Let $A=1^1 +2^2 +3^3 +\cdots+100^{100}.$

$\phi(5^2)=20,(a,5)=1\implies a^{20}\equiv 1\pmod {5^2},a^4\equiv 1\pmod 5.$

$A\equiv (1^1+2^2+3^3+\cdots+20^{20})+(21^1+22^2+\cdots40^{20})+\cdots+(81^1+\cdots+100^{20})\\ \equiv(1+21+41+61+81)+\sum_{k=0}^4\sum_{i=2}^{20}(20k+i)^i \\\equiv 5+\sum_{k=0}^4\sum_{i=2}^{20}(i^i+i\cdot20k\cdot i^{i-1}) \\\equiv 5+\sum_{k=0}^4(20k+1)\sum_{i=2}^{20}i^i \\\equiv 5+5\sum_{i=2}^{20}i^i \\\equiv 5\sum_{i=1}^{20}i^i \pmod {25}$

$\sum_{i=1}^{20}i^i\equiv (1+2^2+3^3+4^0)+(1^2+2^3+3^0+4^1)+(1^3+2^0+3^1+4^2)+(1^0+2^1+3^2+4^3) \\\equiv 3+4+1+1=4\pmod5 \\\implies A \equiv 20\not \equiv 0\pmod{25},A\equiv 0\pmod 5$

It's easy to find that $A$ is even, hence $10\mid A,100\nmid A.$ Hence there is only one zero at the end of $A$.