''Hint

Solution. Let $g(x):=f(x)-x^n$. By Lagrange Interpolation, $\begin{eqnarray*}g(x)&=&\sum_{i=1}^ng(i)\prod_{1\leq j\leq n\atop j\neq i}\frac{x-j}{i-j}\\ &=& \sum_{i=1}^n\frac{g(i)}{\prod_{1\leq j\leq n\atop j\neq i}(i-j)}\prod_{1\leq j\leq n\atop j\neq i}(x-j)\\ &=& \sum_{i=1}^n\frac{g(i)(-1)^{n-i}}{(i-1)!(n-i)!}x^{n-1}\prod_{1\leq j\leq n\atop j\neq i}\left(1-\frac{j}{x}\right) \end{eqnarray*}.$ Dividing by $x^{n-1}$ and taking $x\rightarrow\infty$, we obtain $\begin{eqnarray*} 0 = \sum_{i=1}^n\frac{ig(i)(-1)^{n-i}}{\frac{n!}{\binom{n}{i}}} \end{eqnarray*},$ or equivalently, $\begin{eqnarray*} \sum_{i=1}^n\binom{n}{i}(-1)^{n-i}i^{n+1}=\sum_{i=1}^ni\binom{n}{i}(-1)^{n-i}f(i) \end{eqnarray*}.$ It is not hard to prove (using Lagrange interpolation) that the left hand side is $\frac{n(n+1)!}{2}$. If for all $i\in [n]$, $|f(i)|<\frac{n!}{\binom{n}{i}}$, then $\begin{eqnarray*} \frac{n(n+1)!}{2}=\left| \sum_{i=1}^ni\binom{n}{i}(-1)^{n-i}f(i)\right|\leq \sum_{i=1}^ni\binom{n}{i}|f(i)| a contradiction. $\Box$

Can someone explain more to me?

The broken latex fixed, just removed a couple of redundant $'s boxedexe wrote: Solution. Let $g(x):=f(x)-x^n$. By Lagrange Interpolation, \begin{eqnarray*}g(x)&=&\sum_{i=1}^ng(i)\prod_{1\leq j\leq n\atop j\neq i}\frac{x-j}{i-j}\\ &=& \sum_{i=1}^n\frac{g(i)}{\prod_{1\leq j\leq n\atop j\neq i}(i-j)}\prod_{1\leq j\leq n\atop j\neq i}(x-j)\\ &=& \sum_{i=1}^n\frac{g(i)(-1)^{n-i}}{(i-1)!(n-i)!}x^{n-1}\prod_{1\leq j\leq n\atop j\neq i}\left(1-\frac{j}{x}\right) \end{eqnarray*}. Dividing by $x^{n-1}$ and taking $x\rightarrow\infty$, we obtain \begin{eqnarray*} 0 = \sum_{i=1}^n\frac{ig(i)(-1)^{n-i}}{\frac{n!}{\binom{n}{i}}} \end{eqnarray*}, or equivalently, \begin{eqnarray*} \sum_{i=1}^n\binom{n}{i}(-1)^{n-i}i^{n+1}=\sum_{i=1}^ni\binom{n}{i}(-1)^{n-i}f(i) \end{eqnarray*}. It is not hard to prove (using Lagrange interpolation) that the left hand side is $\frac{n(n+1)!}{2}$. If for all $i\in [n]$, $|f(i)|<\frac{n!}{\binom{n}{i}}$, then \begin{eqnarray*} \frac{n(n+1)!}{2}=\left| \sum_{i=1}^ni\binom{n}{i}(-1)^{n-i}f(i)\right|\leq \sum_{i=1}^ni\binom{n}{i}|f(i)|<n!\sum_{i=1}^ni=\frac{n(n+1)!}{2} \end{eqnarray*}, a contradiction. $\Box$

goodar2006 wrote: We define the polynomial $f(x)$ in $\mathbb R[x]$ as follows: $f(x)=x^n+a_{n-2}x^{n-2}+a_{n-3}x^{n-3}+.....+a_1x+a_0$ Prove that there exists an $i$ in the set $\{1,....,n\}$ such that we have $|f(i)|\ge \frac{n!}{\dbinom{n}{i}}$. Another idea. Assume on the contrary it's not true, i.e. $$ \displaystyle |f(i)|< \frac{n!}{\dbinom{n}{i}}\,\,\,\,\,\,\,\,\,(1)$$The idea is to take $ n-1$-th finite difference with step $ 1$ of $ f(x)$ at $ x=1$ , i.e. $ \Delta_1^{n-1}[f](x)$. One the one hand it can be estimated using $ (1)$. On the other hand it equals $ \Delta^{n-1}_1[x^n](1)$, because $\Delta_h^{n-1}P(x)$ is zero for any polynomial $ P$ of degree $ n-2$. These two estimates contradict each other. Details in my blog.