If n integers taken at random are multiplied together, show that the chance that the last digit of the product
is 1, 3, 7, or 9 is $\frac{2^{n}}{5^{n}}$; the chance of its being 2, 4, 6 or 8 is $\frac{4^{n}-2^{n}}{5^{n}}$; of its being 5 is $\frac{5^{n}-4^{n}}{10^{n}}$; and of its being 0 is $\frac{10^{n}-8^{n}-5^{n}+4^{n}}{10^{n}}$.

I am not able to proceed, the word random complicates the question.