Showing Bernstein polynomial is a basis

When you expand them, you see that only one of the Bernstein polynomials has a non-zero constant term, namely $\binom nn(1-x)^n$. So, if$$\sum_{k=0}^n\alpha_kB_{n,k}(x)=0,\tag1$$then $\alpha_0=0$.

Now, there are only two Bernstein polynomials such that the coefficient of $x$ is non-zero, which are $B_{n,0}(x)$ and $B_{n,1}(x)$. But you already know that $\alpha_0=0$. It follows then from $(1)$ that $\alpha_1=0$.

And so on…

Hint: The matrix that expresses the Bernstein polynomials with respect to the canonical monomial basis is triangular with a diagonal of binomial coefficients, and so is invertible.

For instance, when $n=3$, we have $$ \begin{pmatrix} B_{3,0}(x) \\ B_{3,1}(x)\\ B_{3,2}(x) \\ B_{3,3}(x) \end{pmatrix} = \begin{pmatrix} (1-x)^3 \\ 3x(1-x)^2 \\ 3x^2(1-x) \\ x^3 \end{pmatrix} = \begin{pmatrix} 1 & -3 & \hphantom{-}3 & \hphantom{-}1 \\ 0 & \hphantom{-}3 & -6 & \hphantom{-}3 \\ 0 & \hphantom{-}0 & \hphantom{-}3 & -3 \\ 0 & \hphantom{-}0 & \hphantom{-}0 & \hphantom{-}1 \\ \end{pmatrix} \begin{pmatrix} 1 \\ x \\ x^2 \\ x^3 \end{pmatrix} $$ The exact entries in the matrix are not important. The key point is that the $x^k$ factor in $B_{n,k}(x)$ ensures that in the $k$-th row all entries before the diagonal are zero, and so the matrix is triangular.