Yes.

I claim that

**(1)** $\prod\limits_{i=1}^s \left(M-u_i\right) = \prod\limits_{i=1}^{t} \left(m-u_i\right) \cdot \prod\limits_{i=t+1}^{s} \left(M-u_i\right) $ for every $t \in \left\lbrace 0,1,...,s-1\right\rbrace$.

In fact, you can prove **(1)** by induction over $t$. The base case $t = 0$ is tautological. In the induction step, you need to show that

$\prod\limits_{i=1}^{t} \left(m-u_i\right) \cdot \prod\limits_{i=t+1}^{s} \left(M-u_i\right) = \prod\limits_{i=1}^{t-1} \left(m-u_i\right) \cdot \prod\limits_{i=t}^{s} \left(M-u_i\right)$

for every $t \in \left\lbrace 1,2,...,s-1\right\rbrace$. Clearly, this equation rewrites as

$\left(m-u_t\right)\left(M-u_s\right)A = \left(M-u_t\right)\left(M-u_s\right)A$,

where $A = \prod\limits_{i=1}^{t-1} \left(m-u_i\right) \cdot \prod\limits_{i=t+1}^{s-1} \left(M-u_i\right)$.

Thus, the induction will be complete once we show that

$\left(m-u_t\right)\left(M-u_s\right) = \left(M-u_t\right)\left(M-u_s\right)$.

But this is clear since

$\left(m-u_t\right)\left(M-u_s\right) - \left(M-u_t\right)\left(M-u_s\right)
= -\underbrace{\left(M-m\right)\left(M-u_s\right)}_{=0} = 0$.

So we have proven **(1)** by induction. Applying **(1)** to $t=s-1$, we obtain

$\prod\limits_{i=1}^s \left(M-u_i\right) = \underbrace{\prod\limits_{i=1}^{s-1} \left(m-u_i\right)}_{=\left(m-u_1\right)\left(m-u_2\right)...\left(m-u_{s-1}\right)=0} \cdot \prod\limits_{i=s}^{s} \left(M-u_i\right) = 0$,

qed.