There was another problem very similar to this in hso a few days ago. Write this as $(5+\sqrt{35})^{2n-1} + (5 - \sqrt{35})^{2n-1}$ which is the closed form for the recurrence $f(k) = 10f(k-1)+10f(k-2),$ with initial conditions $f(0) = 2, f(1) = 10.$ and then prove by induction. We want to prove it for $2(k+1) - 1 = 2k+1,$ that $f(2k+1)$ is divisible by $10^{k+1}$ and we have $f(2k+1) = 10f(2k)+10f(2k-1),$ and by induction $f(2k-1)$ is divisible by $10^k,$ so it suffices to show that $f(2k)$ is also divisible by $10^k,$ which we can use induction that $f(2k) = 10f(2k-1)+10f(2k-2),$ and $f(2k-1)$ is divisible by $10^k$ while $f(2k-2)$ is divisible by $10^{k-1},$ so $f(2k)$ is divisible by $10^k$ by induction, and we're done. see here