I think that the $a_i$'s should be integers maybe? Anyways, the below is a solution which assumes that the $a_i$'s are all integral. Let $q = -\frac{b}{a}$ be an arbitrary root of $P$, where $b \ge 0, a> 0$ are relatively prime integers. We know that $q$ is nonpositive due to the fact that all $a_i$'s are non-negative, and $P$ is non-constant. Claim. $ax+b$ divides $P$ in $\mathbb{Z}[x]$. Proof. Let $Q(x) \in \mathbb{Q}[x]$ be so that $Q(x) (ax+b) = P(x).$ We wish to show that $Q \in \mathbb{Z}[x]$ too. If not, then let $c$ be the smallest positive integer $>1$ for which $cQ \in \mathbb{Z}[x]$, and notice that we now know that $cQ(x) (ax+b) = cP(x),$ with $ax+b$ and $cQ$ both being primitive (coefficients relatively prime) and $cP$ not being primitive ($c$ divides all coefficients). This contradicts Gauss' Lemma, and so we actually have $Q \in \mathbb{Z}[x],$ as desired. $\blacksquare$ As a result of the above, we've shown that any such polynomial $P$ must be factorable into a product of linear binomials with nonnegative integer coefficients. Hence, it suffices only to solve the problem for linears by simply taking a linear factor of $P$ instead. Let's consider the polynomial $P(x) = ax + b$, where $a \in \mathbb{N}$ and $b \ge 0$ are relatively prime. We've that $$P(m) P(m+1) P(m+2) \cdots P(n) \ge a^{n-m+1} \frac{n!}{(m-1)!}.$$ Hence, it suffices to show that the "overcounted" part of the above product is at most $a^{n-m+1} (n-m)!.$ Explanation of what "overcounted" means: If $4, 8, 12$ are our numbers, then the one which is divisible by the highest power of $2$ is $8$, so that $2^{v_2(4)} \cdot 2^{v_2(12)}$ is overcounted in the product above, when we want to find the least common multiple. Repeat this for all primes and multiply to get our overcounted part. As a result of the above, it would suffice to show that the overcounted part actually divides $a^{n-m+1} (n-m)!$. In fact, this is very easily proved with simply $v_p$ calculations for all primes, and so we're done. $\square$

Pathological wrote: I think that the $a_i$'s should be integers maybe? We know that $q$ is nonpositive due to the fact that all $a_i$'s are non-negative $ I think you indeed used that they are non negative

Thats a cool problem: WLOG assume that $P$ is a primitive polynomial. Note that, due to Gauss Lemma, since $P$ is reducible in $\mathbb Q[x]$, so it is reducible in $\mathbb Z[x]$. Thus, $P(x) = \prod_{k=1}^{d} (p_k x+q_k)$ where $(p_k, q_k)=1$. Note that, it suffice to consider each linear factor separately or basically, WLOG assume that $P(x)=px+q$ where $(p, q)=1$. Notice that: $$L = \text{lcm} (pm+q, p(m+1)+q, \cdots, pn+q) = \prod p_k^{e_k}$$$$ P = (pm+q)(p(m+1)+q) \cdots (pn+q) = \prod p_k^{g_k}.$$Notice that: $e_k = \nu_{p_k}(ps+q)$ for some $m \le s \le n$ and $t_k = \nu_{p_k}(pk+q)$ for some $t_k \le e_k$ for all $m \le k \le n$. Notice that: $p_k^{t_k}|(ps+q)-(pk+q) = p(s-k)$. Note that: $\gcd(p, p_k)=1$ which implies $p_k^{t_k}|s-k$. Therefore: $p_k^{g_k} | p^{e_k} \prod_{r=n, r \neq s }^{m} (s-r)$. Note that, we can write the product in the form $|\prod_{r=n, r \neq s }^{m} (s-r)| = u! v!$ where $u+v=n-m$. Thus: $|\prod_{r=n, r \neq s }^{m} (s-r)| = u! v! | (n-m)!$. Therefore, we have: $p_k^{g_k} | p^{e_k} (n-m)!$ Taking into product: $P | L (n-m)!$. Thus: $$L (n-m)! \ge P \ge m\frac{n!}{m!}$$$$\implies L \ge n \binom{n}{m}$$and we are done.