do there exist $d_i \in \{ 0,1,2,3,..15\}, p \in \mathbb N$
such that $$10^{-1}= d_1 16^{-1}+d_2 16^{-2}+...+d_p 16^{-p}?$$
my approach : $10^{-1}= (d_1 16^{p-1}+d_2 16^{p-2}+...+d_p)\cdot 16^{-p} \Leftrightarrow 16^{p}=10\cdot (d_1 16^{p-1}+d_2 16^{p-2}+...+d_p)$
But on the left side we have a number ending in $6$ while in the right side we have a number that ends in $0$ , so the two can't be equal. Is this a valid reasoning? Also what would be the "standard" way for solving such a question?
That is good reasoning. I think you are correct.
I think the "standard" argument would be that the right side is a rational number with a denominator divides some power of 16, and no such fraction is equal to $\frac1{10}$ because 10 does not divide $16^{n}$, since it contains a factor of 5, while $16^n$ does not.
But this is quite similar to your argument.