Coefficient of $x^{r}$ I was willing to find the coefficient of $x^{49}$ in the expression $(x+1)(x+2)...(x+100)$
But is there any kind of generalisation of finding the coefficient of $x^r , 0<r<n$ in the expression
$$\prod_{i=1}^{n}(x+i)$$
Did there exist closed form ?
If not 
Then answer my former question only .
Thanks in advance. I feel i will get an answer here surely. I asked it on brilliant.org but no one replied there.
 A: In general you have $\prod^n_i (x - \lambda_i ) = x^n - e_1(\lambda_1,\dots,\lambda_n) \cdot x^{n-1} + \dots + (-1)^{n} e_n(\lambda_1,\dots,\lambda_n)$
where $e_k(X_1,\dots,X_n) = \sum_{1\le i_1 < \dots < i_k \le n} X_{i_1} \cdots X_{i_k}$.
Edit: knowing that $x^{\underline{n}} = x(x-1)\dots(x-(n+1)) = \sum_{k=0}^n (-1)^{n-k} s_{n,k} \cdot x^k$, (where $s_{n,k}$ are the stirling numbers of the first kind) you get a connection to the elementary symmetric polynomials: 
coeff. of $x^{n-k}$ $ = (-1)^k e_k(0,\dots,n-1) = (-1)^k s_{n,n-k}$
A: The coefficient of $x^{49}$ is quite large.  I get 
$$
440203875867633250993831537954997438935190133369140390700 \\
030010603623221873719241373833402230030817400306028361250
$$
which has $114$ digits.  Offhand I don't see a quick way of obtaining the coefficients you want; technically, you can get closed forms, but they're not trivial to compute by hand.  Computer aid makes it trivial, of course, but I'm not sure that's what you're after...
A: There is no closed formula available which could represent the coefficient of $x^r$ in
\begin{align*}
  \prod_{k=1}^n(x+k)
  \end{align*}
But, as already indicated in a comment we can find a representation of the product using Stirling numbers of the first kind $(-1)^{n-k}\left[\begin{array}{c} n \\ k \end{array}\right]$.

We start with close relatives of Stirling numbers, called unsigned Lah numbers $L(n,k)$.
  \begin{align*}
  L(n,k)=\binom{n-1}{k-1}\frac{n!}{k!}\qquad\qquad k,n>0
  \end{align*}
With these numbers we can write rising factorials $x^{\overline{n}}$ in terms of falling factorials $x^{\underline{n}}$ (and vice versa).
  \begin{align*}
  x^{\overline{n}}=x(x+1)\cdots (x+n-1)=\sum_{k=1}^nL(n,k)x^{\underline{k}}\qquad \qquad n>0
  \end{align*}
The falling factorials $x^{\underline{n}}$ can be expressed in terms of Stirling numbers of the first kind $(-1)^{n-k}\left[\begin{array}{c} n \\ k \end{array}\right]$.
  \begin{align*}
  x^{\underline{n}}=x(x-1)\cdots (x-n+1)=\sum_{k=1}^n(-1)^{n-k}\left[\begin{array}{c} n \\ k \end{array}\right]x^k\qquad\qquad n>0
  \end{align*}

Putting all together we obtain a generating function for rising factorials:

\begin{align*}
  x^{\overline{n}}&=x(x+1)\cdots(x+n-1)\\
&=\sum_{k=1}^nL(n,k)\sum_{j=1}(-1)^{j-k}\left[\begin{array}{c} n \\ k \end{array}\right]x^j\\
  &=\sum_{k=1}^n\binom{n-1}{k-1}\frac{n!}{k!}
  \sum_{j=1}^k(-1)^{k-j}\left[\begin{array}{c} k \\ j \end{array}\right]x^j\tag{1}\\
 &=\sum_{j=1}^n\sum_{k=j}^n\binom{n-1}{k-1}\frac{n!}{k!}(-1)^{n-k}
  \left[\begin{array}{c} k \\ j \end{array}\right]x^j\tag{2}
  \end{align*}

Comment:


*

*In (1) we write the unsigned Lah numbers in terms of binomial coefficients

*In (2) we exchange the sums

It's convenient to use the coefficient of operator to denote the coefficient of $[x^n]$ in a series. This way we can write the coefficient of $x^r$ in OPs product using (2) as
\begin{align*}
[x^r](x+1)\cdots (x+n)&=[x^{r+1}]x(x+1)\cdots (x+n)\\
&=[x^{r+1}]x^{\underline{n+1}}\\
&=\sum_{k=r+1}^{n+1}\binom{n}{k-1}\frac{(n+1)!}{k!}(-1)^{n+1-k}
  \left[\begin{array}{c} k \\ r+1 \end{array}\right]\qquad\qquad 0<r\leq n\\
\end{align*}
The coefficient of $x^{49}$ in OPs product can be calculated as
  \begin{align*}
  [x^{49}]&(x+1)\cdots (x+100)=\sum_{k=50}^{101}\binom{100}{k-1}\frac{101!}{k!}(-1)^{k-1}\left[\begin{array}{c} k \\50 \end{array}\right]
  \end{align*}

With some  help of Wolfram Alpha we see the sum coincides with the result of @BrianTung.
Note: We could also represent the Stirling numbers of the first kind in terms of binomial coefficients. See this related question in MO.
