Basis of a polynomial ring over the field of Rationals

If $n\in\Bbb N$,$$x^n=\bigl((x-1)+1\bigr)^n=\sum_{k=0}^n\binom nk(x-1)^k,$$and therefore $x^n$ is a linear combination of polynomials of the form $(x-1)^k$. So, since the $x^n$'s span $\Bbb Q[x]$, then $(x-1)^n$'s also span it.

Now, let us see that $\{1,x-1,(x-1)^2,\ldots,(x-1)^n\}$ is linearly dependent. Let $a_0,a_1,\ldots,a_n$ be scalars such that$$a_0+a_1(x-1)+a_2(x-1)^2+\cdots+a_n(x-1)^n=0.$$If you expand the LHS of the previous equality, you get a polynomial of degree $n$ such that the coefficient of $x^n$ is $a_n$. But the $x^n$'s are linearly independent, and therefore $a_n=0$. So, you know that$$a_0+a_1(x-1)+a_2(x-1)^2+\cdots+a_{n-1}(x-1)^{n-1}=0,$$and you can start all over again, showing that $a_{n-1}=a_{n-1}=\cdots=a_1=a_0=0$.