Schur function principal specialisation Let $s_{\lambda}(p_1,p_2,)$ denote schur function in power-sum symetric basis. More precisely 
$$  s_{\lambda}=\sum_{\mu}\chi_{\mu}^{\lambda}p_{\mu}|k_{\mu}|.(*)$$
$p_i=\sum_k x_k^i$. 
For more notation and clarity please follow 
http://users.math.msu.edu/users/sagan/Papers/Old/schur.pdf
My question as follows if we take the principal specialization that replaces $p_k$ by $x^k$ in the equation (*) we get 
\begin{equation}
\begin{split}
s_{\lambda}(x,x^2,\ldots)& \neq 0  if \lambda\qquad  \text{is trivial partition}\\
                         & = 0 \qquad otherwise
\end{split}
\end{equation}
The above equation is a well-known fact and the proof is not difficult. Now if I replace $p_k$ with $sx^k$ where $s$ is another determinant I get the following form 
\begin{equation}
s_{\lambda}(sx,sx^2,sx^3,\ldots)=x^{|\lambda|}\prod_{\Box\in \lambda}\frac{s+c(\Box)}{h(\Box)}
\end{equation}
where $c(\Box)$ denote the content of the tableaux $\lambda$ and $h(\Box)$ denote the hook length of each  box. \
Can anyone please cite a proof?  
 A: The backbone of the prove based on the fact named in literature as {\textbf{Stanley Hook-content formula }}[\cite{MR1676282},Cor. 7.21.4] which states that number of semi-standard young tableaux of shape $\lambda$ obtained by 
filling the tableaux with positive integer $i$ such that $1\leq i \leq s $ is given by 
 $$ \prod_{\Box \in \Lambda} \frac{s+c(\Box)}{h(\Box)}. $$
 Given a partition $\lambda$ the associated Schur function is defined as 
 $$ s_{\lambda}({\bf{x}})=\sum_{T}x^{T}    $$
for more detail notation we refer to Bruce sage \cite{} section 4.4. Now if we place $x_{1}=1,\ldots,x_{s}=1$ and $x_{k}=0\forall k>s$, then using Stanley Hook-content formula we get 
$$  s_{\lambda}(x_{1},\ldots)|_{x_{i}=x} =  \prod_{\Box \in \Lambda} \frac{s+c(\Box)}{h(\Box)}. $$
Now using the power sum symmetric basis denoted by $p_{i}$ schur function are written as following.
\begin{equation}  s_{\lambda}(p_{1}\ldots)=\frac{1}{n!}\sum_{\mu}\chi_{\mu}^{\lambda}p_{\mu}|k_{\mu}|
\end{equation}
where $|k_{\mu}|$ denote the cardinality of conjugacy class corresponding to $\mu$. $p_{\mu}=\prod_{i} p_{\mu_{i}}.$
Substituting $p_{i}=sx^{i}$ in the eq(1) bring out a factor $x^{|\lambda |}$ as all are partition of fixed integer. So the R.H.S of eq(1) become following 
$$x^{|\lambda |} s_{\lambda}(p_{1},\ldots)|_{p_{i}=s}.$$ 
Solving the equation $p_{i}=s$ for all $i$, gives us $x_{j}=1$  for $1\leq j \leq s$ and $x_{j}=0 , j > s$, hence our proposition follows from the very first discussion of the proof. 
