Proof of the determinant of the Vandermonde matrix via induction Let $V_n$ be the Vandermonde matrix made up of a sequence of $n$ real numbers $\lambda_1, ..., \lambda_n$,
\begin{align*}
    \begin{pmatrix}
        1 & \ldots & 1 \\ 
        \lambda_1 & \ldots & \lambda_n \\ 
        \vdots & \ddots & \vdots \\ 
        \lambda_1^{n-1} & \ldots & \lambda_n^{n-1}
    \end{pmatrix} 
.\end{align*}
I was requested to prove via induction that
\begin{align*}
    \det(V_{n}) &= \prod_{1\leq i \leq j \leq n}^{n} (\lambda_j -\lambda_i)
\end{align*}
I only started to study the concept of determinant yesterday, and after solving a few simple problems I encountered this one and got stuck. In any case, keep in mind that I'm a neophyte in the subject. Here's what I attempted.
$I$. The base case is trivial. Let us assume for an arbitrary $k-1$ that
\begin{align*}
    \det(V_{k-1}) &= \prod_{1\leq i \leq j \leq n}^{k-1} (\lambda_j -\lambda_i)
\end{align*}
and let us observe the case for $n=k$. The only thing I could think of was to use the recursive definition of the determinant, observing that
\begin{align*}
    \det(V_k) &= \sum_{i=1}^k (-1)^{k+1}v_{i1} \det(V_k(i|1)) \\ 
              &= \det(V_{k-1}) + (-1)^{k+1}v_{k1}\det(V_k(k|1)) \\ 
              &= \prod_{1\leq i \leq j \leq n}^{k-1} (\lambda_j -\lambda_i) +
              (-1)^{k+1}\lambda_1^{k-1}\det(V_k(1|k))
.\end{align*}
Here, $A(i|j)$ is a function that maps some matrix $A$ to a matrix that is equal to $A$ except that the $i$th row, $j$th column were removed. So if $A$ has dimension $n^2$, $A(i|j)$ has dimension $(n-1)^2$. Observe then that $V_k(k|1)$ is a $(k-1) \times (k-1)$ Vandermonde matrix made of the sequence $\lambda_2, ..., \lambda_k$. Therefore our inductive hypothesis applies to it. Then
\begin{align*}
    \det(V_k) &= \prod_{1\leq i \leq j \leq n}^{k-1} (\lambda_j -\lambda_i) +
    (-1)^{k+1} \lambda_1^{k-1}\prod_{2\leq i \leq j \leq n}^{k} (\lambda_j -\lambda_i)
.\end{align*}
I was unable to advance from here. The fact that this is a sum makes it complicated to reduce the whole expression to a single $\prod$ multiplication. I think this approach, via the recursive definition of the determinant, is not ideal, but I can't think of alternatives.
Thanks in advance.
 A: In your formulas you should write $i<j$ as the first commenter has pointed out a few days ago. Otherwise, all your products will be zero. There is also a lot of confusion with $n$ and $k$ in your formulas.
Expanding the determinant by the last row gives
$$\tag{1}
\det(V_n)=\sum_{k=1}^n (-1)^{n+k}\,\lambda_k^{n-1}\det(V_{\hat{k},n-1})
$$
where $V_{\hat{k},n-1}$ is the Vandermonde matrix of
$\lambda_1,\dots\,\lambda_{k-1},\lambda_{k+1},\dots,\lambda_n$ to which the
induction assumption applies:
$$\tag{2}
\det(V_{\hat{k},n-1})=\prod_{\textstyle\genfrac{}{}{0pt}{}{1\leq i < j \leq n}{i,j\not=k}} (\lambda_j -\lambda_i)\,.
$$

*

*Without loss of generality we can assume $\lambda_1<\dots<\lambda_n$ since
every sorting of the $\lambda_i$ (i.e. the columns of $V_n$) will at most only change the sign of the determinant. By (2) the sign of $\det(V_{\hat{k},n-1})$ is then positive.


*Since $(-1)^{n+n}=1$ we see from (1) that $\det(V_n)$ is a polynomial in $\lambda_n$ of degree $n-1$ with leading coefficient one.


*When two $\lambda_j$ and $\lambda_i$ are equal then $\det(V_n)$ has two equal
columns and must be zero. Therefore, the polynomial $\det(V_n)$ has the roots
$\lambda_1,\dots,\lambda_{n-1}\,.$ Since it is of degree $n-1$ it must factorize into the form
$$\tag{3}
f(\lambda_1,\dots,\lambda_{n-1})(\lambda_n-\lambda_{n-1})\dots(\lambda_n-\lambda_1)
$$
where $f$ is a function that cannot depend on $\lambda_n\,.$


*Likewise, $\det(V_n)$ must be a polynomial of degree $n-1$ in $\lambda_{n-1}$
with zeroes $\lambda_1,\dots,\lambda_{n-2}$ and $\lambda_n\,.$ Since
the factor $(\lambda_n-\lambda_{n-1})$ is already isolated in (3) we conclude that
$f$ is of the form
$$\tag{4}
g(\lambda_1,\dots,\lambda_{n-2})(\lambda_{n-1}-\lambda_{n-2})\dots(\lambda_{n-1}-\lambda_1)
$$
where $g$ does not depend on $\lambda_{n-1}$ and $\lambda_n\,.$


*Applying this reasoning to all other $\lambda_i$ shows that $\det(V_n)$ is of the form
$$\tag{5}
\det(V_n)=c\cdot\prod_{1\leq i < j \leq n} (\lambda_j -\lambda_i)
$$
where $c$ is only a constant. It remains to notice  that $c=1\,.$ Indeed,
by expanding (5) is is identified as
the coefficient of $\lambda_n^{n-1}$ which was seen to be one
in the second bullet point above.
