$$S_i = \int^1_0 p_1(x_1) \int^1_{x_1} p_2(x_2) ... \int^1_{x_{i-1}} p_i(x_i) dx_i dx_{i-1}...dx_1$$
where $p_j(z) = \lambda_j e^{-\lambda_j z}$
So I tried just going through with it, but it just turns out to be a huge mess. I was thinking of coming up with some kind of a recursive formula.
$$S_i = \text{Some messy integral} - S_{i-1}$$ My calculations led me to this, but I couldn't see a pattern which could help me find its value.
If I am not mistaken:
\begin{align} P_i(x_{i-1})&:=\int_{x_{i-1}}^1p_i(x_i)\,dx_i=e^{-\lambda_i x_{i-1}}-e^{-\lambda_i}\,,\\ P_{i-1}(x_{i-2})&:=\int_{x_{i-2}}^1p_{i-1}(x_{i-1})P_i(x_{i-1})\,dx_{i-1} =\underbrace{\lambda_{i-1}\,e^{-\lambda_i}\frac{1-\lambda_i-\lambda_{i-1}}{\lambda_i+\lambda_{i-1}}}_{=:\Lambda_{i-1}}P_i(x_{i-2})\,,\\ P_{i-2}(x_{i-3})&:=\int_{x_{i-3}}^1p_{i-2}(x_{i-2})P_{i-1}(x_{i-2})\,dx_{i-2}=\Lambda_{i-1}\,\underbrace{\lambda_{i-2}\,e^{-\lambda_i}\frac{1-\lambda_i-\lambda_{i-2}}{\lambda_i+\lambda_{i-2}}}_{=:\Lambda_{i-2}}P_i(x_{i-3})\,. \end{align} Therefore, we should have $$ S_i=\Big(\prod_{j=1}^{i-1}\Lambda_j\Big)P_i(0)=\Big(\prod_{j=1}^{i-1}\Lambda_j\Big)(1-e^{-\lambda_i})\,. $$