solving determinant with different variable on diagonal but same variable elsewhere? I'm trying to solve a problem from a textbook. This is a listed problem under the first chapter of the textbook (the book starts with introducing determinant without matrix theory) so the only things that are allowed to use are properties of determinant and calculus.

Problem: show that
$$D=\begin{vmatrix}a_1&\lambda&\lambda&...&\lambda\\\lambda&a_2&\lambda&...&\lambda\\\lambda&\lambda&a_3&...&\lambda\\...&...&...&...&...\\\lambda&\lambda&\lambda&...&a_n\end{vmatrix}=\varphi(\lambda)-\lambda\frac{d\varphi}{d\lambda}$$
where $\varphi(\lambda)=(a_1-\lambda)(a_2-\lambda)\ ...\ (a_n-\lambda)$
 A: Edit: after checking through my steps several times, I got the answer:
[Step 1]: Let $a_1=\lambda(1+\sigma_1),\ a_2=\lambda(1+\sigma_2),\ ...,\ a_n=\lambda(1+\sigma_n)$
$$D=\begin{vmatrix}\lambda(1+\sigma_1)&\lambda&\lambda&...&\lambda\\\lambda&\lambda(1+\sigma_2)&\lambda&...&\lambda\\\lambda&\lambda&\lambda(1+\sigma_3)&...&\lambda\\...&...&...&...&...\\\lambda&\lambda&\lambda&...&\lambda(1+\sigma_n)\end{vmatrix}\\=\lambda^n\begin{vmatrix}(1+\sigma_1)&1&1&...&1\\1&(1+\sigma_2)&1&...&1\\1&1&(1+\sigma_3)&...&1\\...&...&...&...&...\\1&1&1&...&(1+\sigma_n)\end{vmatrix}\\=\lambda^n\begin{vmatrix}(1+\sigma_1)&1&1&...&1\\-\sigma_1&\sigma_2&0&...&0\\-\sigma_1&0&\sigma_3&...&0\\...&...&...&...&...\\-\sigma_1&0&0&...&\sigma_n\end{vmatrix}\\=-\sigma_1\lambda^n\begin{vmatrix}\frac{(1+\sigma_1)}{-\sigma_1}&1&1&...&1\\1&\sigma_2&0&...&0\\1&0&\sigma_3&...&0\\...&...&...&...&...\\1&0&0&...&\sigma_n\end{vmatrix}$$
[Step 2]: apply Laplace expansion for the 1st column
$$D=-\sigma_1\lambda^n[(-1)^{1+1}(\sigma_2\sigma_3...\sigma_n)\frac{(1+\sigma_1)}{-\sigma_1}+(-1)^{1+2}(-1)^{1+1}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_2}+(-1)^{1+3}(-1)^{1+2}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_3}+(-1)^{1+4}(-1)^{1+3}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_4}+...+(-1)^{1+n}(-1)^{n}\frac{(\sigma_2\sigma_3...\sigma_n)}{\sigma_n}]\\=\lambda^n(1+\frac{1}{\sigma_1}+\frac{1}{\sigma_2}+\frac{1}{\sigma_3}+...+\frac{1}{\sigma_n})(\sigma_1\sigma_2\sigma_3...\sigma_n)$$
[Step 3]: Because $a_1=\lambda(1+\sigma_1),\ a_2=\lambda(1+\sigma_2),\ ...,\ a_n=\lambda(1+\sigma_n)$
$$\varphi(\lambda)=\lambda^n(\sigma_1\sigma_2...\sigma_n)\ \ \ \textbf{[1]}$$
$$\lambda\frac{d\varphi(\lambda)}{d\lambda}=\lambda\frac{d}{d\lambda}e^{ln(\varphi(\lambda))}=\lambda(\frac{1}{\lambda-a_1}+\frac{1}{\lambda-a_1}+...+\frac{1}{\lambda-a_n})\varphi(\lambda)=(-\frac{1}{\sigma_1}-\frac{1}{\sigma_2}-...-\frac{1}{\sigma_n})\varphi(\lambda)\ \ \ \textbf{[2]}$$
$$\varphi(\lambda)-\lambda\frac{d\varphi}{d\lambda}=\lambda^n(1+\frac{1}{\sigma_1}+\frac{1}{\sigma_2}+...+\frac{1}{\sigma_n})(\sigma_1\sigma_2...\sigma_n)\ \ \ \textbf{[3]}$$
