The first digit (on the left) and the last digit (on the right) of $$\sum_{k=1}^{1010}k^{2020-k}=1^{2019}+2^{2018}+3^{2017}+\dots+1010^{1010}$$ are to be obtained not by using computers/calculators.
My Attempt for the last digit (the units digits):
$$\left.\begin{matrix} 1^{2019}\equiv 1 \text{ (mod 10)}\\ 2^{2018}\equiv 4 \text{ (mod 10)}\\ 3^{2017}\equiv 3 \text{ (mod 10)}\\ 4^{2016}\equiv 6 \text{ (mod 10)}\\ 5^{2015}\equiv 5 \text{ (mod 10)}\\ 6^{2014}\equiv 6 \text{ (mod 10)}\\ 7^{2013}\equiv 7 \text{ (mod 10)}\\ 8^{2012}\equiv 6 \text{ (mod 10)}\\ 9^{2011}\equiv 9 \text{ (mod 10)}\\ 10^{2010}\equiv 0 \text{ (mod 10)}\\ \end{matrix}\right\} \text{ this pattern repeats } \frac{1010}{10}=101 \text{ times for the next terms.}$$
The units digit of the sum of the first $10$ terms
$=(1+4+3+6+5+6+7+6+9+0)$ $($mod $10)=47$ $($mod $10)=7$
Therefore, the units digit of the given expression is $(7 \times 101)$ $($mod $10)=707$ $($mod $10)=7$.
I am not sure if I am right or wrong. Please let me know.
Also, I do not know anyway to find the first digit (on the left).
Any help would be appreciated. THANKS!