 Solution

Suppose ${\sum }_{i=1}^{k+1}{\alpha }_i{u}_i=0$. If ${\alpha }_{k+1}\ne 0$, thencontrary to the hypothesis. Hence ${\alpha }_{k+1}=0$. The assumed linear independence of $u_1,\dots ,u_k$ then gives ${\alpha }_1=\cdots ={\alpha }_k=0$. Therefore $u_1,\dots ,u_{k+1}$ are linearly independent. Adding a vector outside the previous linear span creates no nontrivial relation.