[proofplan]
We evaluate the bialternant formula for $s_\lambda$ at the geometric progression $x_r = q^{r-1}$. Both numerator and denominator become alternant determinants, and after reversing columns the same reversal sign occurs in both determinants, so it cancels in the quotient. The remaining determinants are ordinary Vandermonde products. Factoring each numerator and denominator factor then isolates the total power $q^{\sum_{i=1}^n (i-1)\lambda_i}$ and leaves the stated product.
[/proofplan]
[step:Evaluate the bialternant formula at the geometric progression]
Let $K := \mathbb{Q}(q)$ denote the rational function field in the indeterminate $q$. For $1 \le r \le n$, define
\begin{align*}
x_r := q^{r-1} \in K.
\end{align*}
The bialternant formula for the Schur polynomial $s_\lambda$ in $n$ variables gives
\begin{align*}
s_\lambda(x_1,\dots,x_n)
&=
\frac{
\det\left(x_r^{\lambda_c+n-c}\right)_{1 \le r,c \le n}
}{
\det\left(x_r^{n-c}\right)_{1 \le r,c \le n}
}.
\end{align*}
Substituting $x_r=q^{r-1}$ gives
\begin{align*}
s_\lambda(1,q,\dots,q^{n-1})
&=
\frac{
N
}{
D
},
\end{align*}
where
\begin{align*}
N &:= \det\left(q^{(r-1)(\lambda_c+n-c)}\right)_{1 \le r,c \le n},\\
D &:= \det\left(q^{(r-1)(n-c)}\right)_{1 \le r,c \le n}.
\end{align*}
These determinants are elements of $K$, and $D \ne 0$ in $K$ because its Vandermonde product computed below is nonzero.
[/step]
[step:Convert both alternants into Vandermonde determinants]
Let $\rho: \{1,\dots,n\} \to \{1,\dots,n\}$ be the reversing permutation defined by $\rho(c)=n+1-c$. Its sign is
\begin{align*}
\operatorname{sgn}(\rho)=(-1)^{n(n-1)/2}.
\end{align*}
Define elements $y_c,z_c \in K$ for $1 \le c \le n$ by
\begin{align*}
y_c &:= q^{\lambda_{n+1-c}+c-1},\\
z_c &:= q^{c-1}.
\end{align*}
Reversing the columns of $N$ and $D$ gives
\begin{align*}
N &= \operatorname{sgn}(\rho)\det\left(y_c^{r-1}\right)_{1 \le r,c \le n},\\
D &= \operatorname{sgn}(\rho)\det\left(z_c^{r-1}\right)_{1 \le r,c \le n}.
\end{align*}
For any elements $u_1,\dots,u_n$ in a commutative ring, the Vandermonde determinant identity is
\begin{align*}
\det\left(u_c^{r-1}\right)_{1 \le r,c \le n}
=
\prod_{1 \le c < d \le n}(u_d-u_c).
\end{align*}
Applying this identity to $(y_c)_{c=1}^n$ and $(z_c)_{c=1}^n$, and cancelling the identical column-reversal signs, yields
\begin{align*}
s_\lambda(1,q,\dots,q^{n-1})
&=
\prod_{1 \le c < d \le n}
\frac{y_d-y_c}{z_d-z_c}.
\end{align*}
[guided]
The determinant in the bialternant formula is not immediately in the usual Vandermonde shape because the exponents $\lambda_c+n-c$ and $n-c$ decrease with the column index $c$. The standard Vandermonde matrix has columns indexed by bases $u_c$ and rows indexed by powers $0,1,\dots,n-1$, namely $\left(u_c^{r-1}\right)_{r,c}$.
We therefore reverse the columns. Let $\rho:\{1,\dots,n\}\to\{1,\dots,n\}$ be the reversing permutation $\rho(c)=n+1-c$. Reversing $n$ columns has sign
\begin{align*}
\operatorname{sgn}(\rho)=(-1)^{n(n-1)/2}.
\end{align*}
For the numerator, after replacing the original column index by $n+1-c$, the exponent becomes
\begin{align*}
\lambda_{n+1-c}+n-(n+1-c)=\lambda_{n+1-c}+c-1.
\end{align*}
Thus, if
\begin{align*}
y_c := q^{\lambda_{n+1-c}+c-1},
\end{align*}
then the reversed numerator determinant is
\begin{align*}
\det\left(y_c^{r-1}\right)_{1 \le r,c \le n}.
\end{align*}
Including the sign of the column reversal,
\begin{align*}
N = \operatorname{sgn}(\rho)\det\left(y_c^{r-1}\right)_{1 \le r,c \le n}.
\end{align*}
The same column reversal applies to the denominator. If
\begin{align*}
z_c := q^{c-1},
\end{align*}
then
\begin{align*}
D = \operatorname{sgn}(\rho)\det\left(z_c^{r-1}\right)_{1 \le r,c \le n}.
\end{align*}
The two signs are identical because the same reversal permutation is used in both determinants. Hence the signs cancel in the quotient.
Now use the Vandermonde determinant identity
\begin{align*}
\det\left(u_c^{r-1}\right)_{1 \le r,c \le n}
=
\prod_{1 \le c < d \le n}(u_d-u_c).
\end{align*}
Applying it first to $u_c=y_c$ and then to $u_c=z_c$ gives
\begin{align*}
\frac{N}{D}
=
\prod_{1 \le c < d \le n}
\frac{y_d-y_c}{z_d-z_c}.
\end{align*}
This is the key reduction: the Schur specialization is now a quotient of explicit pairwise factors.
[/guided]
[/step]
[step:Relabel the Vandermonde factors by the parts of the partition]
For a pair $1 \le c < d \le n$, define indices
\begin{align*}
i := n+1-d,\qquad j := n+1-c.
\end{align*}
Then $1 \le i < j \le n$, and the correspondence $(c,d)\mapsto(i,j)$ is a bijection from pairs $c<d$ to pairs $i<j$. Under this relabelling,
\begin{align*}
y_d &= q^{\lambda_i+n-i},&
y_c &= q^{\lambda_j+n-j},\\
z_d &= q^{n-i},&
z_c &= q^{n-j}.
\end{align*}
Therefore
\begin{align*}
s_\lambda(1,q,\dots,q^{n-1})
&=
\prod_{1 \le i < j \le n}
\frac{
q^{\lambda_i+n-i}-q^{\lambda_j+n-j}
}{
q^{n-i}-q^{n-j}
}.
\end{align*}
[/step]
[step:Extract the common powers from each factor]
Fix a pair $1 \le i < j \le n$. Since
\begin{align*}
\lambda_i+n-i
=
\lambda_j+n-j+\lambda_i-\lambda_j+j-i,
\end{align*}
we factor the numerator as
\begin{align*}
q^{\lambda_i+n-i}-q^{\lambda_j+n-j}
&=
q^{\lambda_j+n-j}
\left(q^{\lambda_i-\lambda_j+j-i}-1\right).
\end{align*}
Similarly, since
\begin{align*}
n-i = n-j+j-i,
\end{align*}
we factor the denominator as
\begin{align*}
q^{n-i}-q^{n-j}
&=
q^{n-j}\left(q^{j-i}-1\right).
\end{align*}
Thus each factor satisfies
\begin{align*}
\frac{
q^{\lambda_i+n-i}-q^{\lambda_j+n-j}
}{
q^{n-i}-q^{n-j}
}
&=
q^{\lambda_j}
\frac{q^{\lambda_i-\lambda_j+j-i}-1}{q^{j-i}-1}\\
&=
q^{\lambda_j}
\frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}}.
\end{align*}
Multiplying over all pairs $1 \le i < j \le n$ gives
\begin{align*}
s_\lambda(1,q,\dots,q^{n-1})
&=
\left(\prod_{1 \le i < j \le n}q^{\lambda_j}\right)
\prod_{1 \le i < j \le n}
\frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}}.
\end{align*}
[/step]
[step:Compute the extracted exponent and conclude]
For a fixed index $j \in \{1,\dots,n\}$, the factor $q^{\lambda_j}$ occurs once for each $i$ with $1 \le i < j$, hence exactly $j-1$ times. Therefore
\begin{align*}
\prod_{1 \le i < j \le n}q^{\lambda_j}
&=
q^{\sum_{j=1}^{n}(j-1)\lambda_j}.
\end{align*}
Substituting this into the product formula obtained above yields
\begin{align*}
s_\lambda(1,q,q^2,\dots,q^{n-1})
&=
q^{\sum_{j=1}^{n}(j-1)\lambda_j}
\prod_{1 \le i < j \le n}
\frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}}.
\end{align*}
Renaming the dummy index $j$ in the exponent as $i$ gives exactly
\begin{align*}
s_\lambda(1,q,q^2,\dots,q^{n-1})
&=
q^{\sum_{i=1}^{n}(i-1)\lambda_i}
\prod_{1 \le i < j \le n}
\frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}}.
\end{align*}
This proves the stated identity in $\mathbb{Q}(q)$.
[/step]
