Proving by strong induction that $\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $

Hint $\ $ Any sequence of $\rm\,d\,$ consecutive naturals has an element divisible by $\rm\,d.\,$ This has a very simple proof by induction: shifting such a sequence by one does not change its set of remainders mod $\rm\,d,\,$ since it simply replaces the old least element $\rm\:\color{#C00}n\:$ by the new greatest element $\rm\,\color{#C00}{n+d}$

$$\begin{array}{}\rm \:\color{#C00}n &\rm n+1 &\rm n+2 &\rm\! \cdots\! &\rm n+d-1 & \\ \to &\rm n+1 &\rm n+2\rm &\! \cdots\! &\rm n+d-1 &\rm \color{#C00}{n+d} \end{array}$$

Since $\rm\: \color{#C00}n\equiv \color{#C00}{n+d} \pmod{\! d},\,$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\rm\ 0,1,2,\ldots,d-1\ =\: $ all possible remainders mod $\rm\,d.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\rm\,d.$

Note that $n(n+1)\dots(n+d-1)$ is the product of $d$ consecutive integers. Thus, it suffices to prove that if $n,n+1,\dots,n+d-1$ are any $d$ consecutive integers, then $d$ divides one of these integers. I would prove this by induction on $n$, simultaneously for all $d$.

First, it’s clearly true for $n=1$, since in that case the $d$ integers are $1,2,\dots,d$. Suppose now that it’s true for some $n$, and consider the $d$ consecutive integers $n+1,n+2,\dots,n+d$. By the induction hypothesis one of the integers $n,n+1,\dots,n+d-1$ is a multiple of $d$. If it’s one of the integers $n+1,n+2\dots,n+d-1$, you’re done (why?), and if $n$ is a multiple of $d$, then so is ... ?

Hint

$$ (n+1)(n+2)...(n+d-1)(n+d)= \left[(n+1)(n+2)(n+3)...(n+d-1)\right]n + \left[(n+1)(n+2)(n+3)...(n+d-1) \right]d$$

$P(n)$ tells you that the first term on RHS is divisible by $d$, while the second one is clearly divisible by $d$...

Fixed $d$, it is easy to show that the statement holds for every $n$, by induction. The inductive step is as follows: suppose $d|n(n+1)\cdots(n+d-1)$, then put $m=gcd(d,n)$ and observe that $gcd(d,n+d)=m$ as well, therefore, by assumption $$\frac{d}{m}\vert (n+1)\cdots(n+d-1)$$ so $$d=\frac{d}{m}m\vert (n+1)\cdots(n+d-1)(n+d)\;.$$ Now you just have to show that the statement is true for $n=2$ and every $d$, but this is easy: $2\leq d\leq d+2-1=d+1$, therefore $d$ divides $2\cdot3\cdots(d+1)=(d+1)!$.