Some clarifying questions: Do you mean $d[n].C_{n+j}^n$ as the product of those two terms? The claim is that $u$ divides $d[n]\binom{n+j}{n}$?

Do you mind editing your question to show the proof you attempted?

yes us you thinked above

$c_{p}^h = \displaystyle \frac{p!}{h!(p-h)!}$

ok i will try to write first we write u as decomposition of the prime number product of $ p_i^{alpha_i}$ $ p_i$ two by two distincts. we will proove that each $p_i^{alpha_i}$ divide de second member and the proof will be compete. let's take one of the $p_i$ , call him $p$ for simplicity and $alpha_i=\alpha$. we will distinguish two cases $p> n+1$ and $p \leq n$

we suppose that $p \geq n+1$ in this case we have $p^2 \geq n^2 > 2*n>n+j \geq u $ thus $p^2>u$ thus $\alpha$ must be equal to $1$. (p^2 can't divide u since p^2>u$. now $p^{alpha} =p$ divide $u$ and $u$ is one of the factor(n+1)....(n+j) thus p divide this last product, further more( j! divide this last product because the binomial coefficient is an integer

sorry for above Now $p^{\alpha}=p $ divide $u$ and $u$ is one of the term in the product $(n+1)....(n+j)$

pgcd $(j! , p)=1 $ because $p>j$ and $ p$ is prime thus j! p divide the product $(n+1)...(j+n)$ and we obtein p divide the binomial coefficient and so the second hand

the second case $ p\leq n $ is more complicate and in this cas $d[n]$ will be usefull.

let's call$ v_p(z) $ the power of $ p$ in $z$. we have to proove v_p(d[n]\binom{n+j}{n}) \geq v-p(u)=alpha$.

$v_p(d[n]\binom{n+j}{n}) \geq v-p(u)=alpha$.

sorry $ \geq v_p(u)=\alpha$

Is $n \geq 5$ really important? If it is, it provides a lower bound on how nice the proof can be...

sorry $[ log /log ]$ is integer part , if it's equal to $ \alpha $ it's ok, if not that means equal to $ \alpha-1$ we easily proof that one of the term $ [\displaystyle {n+j}{p^k}]-[\displaystyle \frac{n}{p^k}]-[\displaystyle{j}{p^k}], k \geq 1$ $ wich are always $ =0 or 1$ is equal to 1 (this term coreespondind to $k=\alpha$ ). the proof is complited since v_p( binom(n+j,j)$ is equal to the the sum of the term above ( k \geq 1)

$ \displaystyle[ {n+j}{p^k}]-[\displaystyle \frac{n}{p^k}]-[\displaystyle \frac{j}{p^k}], k \geq 1$( wich are always$=0 or 1$) the proof is complited since $v_p$ of the binomial coefficient is = to the sum of the last term writen above with $ p^k$

I added a simple proof of your divisibility relation. I suggest that you delete your comments. (Note that in this site we do not discuss known proofs. If needed, updated your post, comments serve a different purpose.)

sorry $ [\displaystyle \frac {n+j}{p^k}]-[\displaystyle \frac{n}{p^k}]-[\displaystyle \frac{j}{p^k}],$

$ n \geq 5$ could be replacede by $n \geq 3 $ i think

Your divisibility relation holds for all $n\geq 1$ and $0\leq j\leq n$. See my post.

can't find the argument of divisibility

may be but i'm not interested to case $u =2*n$ because $n $ divide $d[n]$

The left hand side of my second display is the exponent of $p$ of the left hand side of my first display. The right hand side of my second display is the exponent of $p$ of the right hand side of my first display. So my second display is equivalent to my first display. You see, I gave a proof in a few lines, and you should accept my response (so that it turns green).

sorry, do you know how i can erase the comments,i'm knew in this site and i don't know how to do it

If you move your mouse over a comment of yours, you will see a red "Delete" button next to the blue "Edit" button. Just push it.

thanks, i hope you can finish the details of the proof at the second case, (may be i'll write them after because i'm tired), but the steps are nor very difficult, i don't like this proof and i'm searching one more simple

I gave a simple and short full proof in my post below. So I am not sure what else do you want.

$v_p(d[n])=[\displaystyle \frac{log(n)}{log(p)}]$ $ [z]$ is the integer part of a real $z$

$v_p(\binom{n+j}{n})= \sum_{k=1}^{+ \infty} [\displaystyle \frac {n+j}{p^k}]-[\displaystyle \frac{n}{p^k}]-[\displaystyle \frac{j}{p^k}]$

sorry i didn't understand your proof

My first display says that $\mathrm{lcm}(1,2,\dots,n+j)$ divides $\mathrm{lcm}(1,2,\dots,n)\binom{n+j}{j}$. However, your $u$ divides $\mathrm{lcm}(1,2,\dots,n+j)$. So your $u$ divides $\mathrm{lcm}(1,2,\dots,n)\binom{n+j}{j}$.