Given an $n \times n$ vandermonde matrix $V$ which is invertible, is any $(n1) \times (n1)$ submatrix of $V$ invertible also? I think the answer is yes, but I don't know how to prove.
This is an elaboration of Thierry's answer: If $0 \leq a_0 < a_1 < \cdots < a_{d1}$ is any sequence of nonnegative integers, then the determinant $\det \left( x_j^{a_i} \right)$ is equal to $\prod_{i < j}(x_i  x_j) \cdot s_{\lambda}(x_1, \ldots, x_d)$ where $\lambda = (a_0, a_11, a_22, \ldots, a_{d1}  (d1))$ and $s_{\lambda}$ is the Schur function. (Wikipedia takes this as the definition of Schur functions.) In your case, you want $(a_0, a_1, \ldots, a_d)$ to be $(0,1,2,3,\ldots, dk1, dk+1, dk+2, \ldots, d)$. The corresponding partition $\lambda$ is $(0,0,0,\ldots, 1,1,\ldots, 1)$ where there are $k$ ones. This Schur function is the elementary symmetric function $$e_k(x) := \sum_{i_1 < i_2 < \cdots < i_k} x_{i_1} x_{i_2} \cdots x_{i_k},$$ which is apparently also known as a Veite sum. So your minor vanishes precisely when there is a repeated column, or when $e_k$ is $0$. 


Well, you're almost right... Such a determinant is sometimes known as "lacunary vandermonde" determinant, although my internet searches have not given much. Lemma 4 in this paper states that the lacunary vandermonde determinant of variables $(x_1, \dots, x_n)$ missing the terms in $x_i^k$ is obtained by a product of the regular vandermonde of $(x_1, \dots, x_n)$ times a Viete sum. You can work out for yourself how to prove this: it's not very difficult, but it's a good exercise. The point though is that, since you assumed your vandermonde was invertible, your lacunary determinant vanishes iff the Viete sum is zero. So there is a hypersurface of possible values for $(x_1, \dots, x_n)$ that cancel your lacunary determinant. 


No. Take $n=2$ and $\alpha_1=0, \alpha_2=1$. Then the Vandermonde matrix \[ \begin{bmatrix} 1 & 0 \\\ 1 & 1 \end{bmatrix} \] is invertible, but the upperright submatrix is not. 


Recognizing the functions $e_k$ in David's excellent elaboration of Thierry's answer leads to a simple (equivalent) description and a proof for the problem considered here (but probably not the more general one of $n1$ arbitrary powers.) Consider the Vandermonde matrix $$V= \begin{bmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ x_1 & x_2 & x_3 & \ldots & x_{n1} & t \\ x_1^2 & x_2^2 & x_3^2 & \ldots & x_{n1}^2 &t^2 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_1^{n1} & x_2^{n1} & x_3^{n1} & \ldots & x_{n1}^{n1} & t^{n1} \end{bmatrix}$$ Claim: The minor obtained from removing the row and column of $t^{n1k}$ is noninvertible exactly when the $x_i$ are the $n1$ (distinct) roots of a polynomial $$a_0t^{n1}+a_1t^{n2}+\cdots+a_{n2}t+a_{n1}$$ with $a_k=0.$ Sketch: Think of $t$ as a variable and the various $x_j$ as parameters. The determinant is a sum of $n!$ terms each with total degree $\frac{n^2n}{2}$ in $t$ and the $x_j$. It is also a polynomial $f(t)$ with coefficients polynomals in the $x_i$. This determinant is zero if any two of the columns are equal. Hence $f(t)$ is identically zero if any two $x_i$ are equal so $$f(t)=g(t)\prod_{0<i<j<n}(x_jx_i)$$ Where $g(t)=e_0t^{n1}+e_1t^{n2}+\cdots+e_{n2}t+e_{n1}$ and $e_k$ has degree $k$ in the $x_j.$ Observe that
Other notes:
It is worth looking at the three cases of the last row, the first row, and the other rows.
That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the description in terms of the $e_k.$ 

