By Taylor's theorem for each $1 \le k \le n-1$ there exists $y_k \in (x,x+k)$ such that $$\sum_{j=1}^{n-1}\frac{f^{(j)}(x)}{j!}k^j=f(x+k)-f(x)-\frac{f^{(n)}(y_k)}{n!}k^n,$$which can be written as $$\begin{pmatrix}1 & 1 & \cdots & 1 \\ 2 & 2^2 & \cdots & 2^{n-1} \\ . & . & \cdots & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ n-1 & (n-1)^2 & \cdots & (n-1)^{n-1} \end{pmatrix}\begin{pmatrix}f'(x) \\ \frac{f''(x)}{2} \\ . \\ . \\ . \\ \frac{f^{(n-1)}(x)}{(n-1)!}\end{pmatrix}=\begin{pmatrix}g_1(x) \\ g_2(x) \\ . \\ . \\ . \\ g_{n-1}(x)\end{pmatrix},$$where $$g_j(x):=f(x+j)-f(x)-\frac{f^{(n)}(y_j)}{n!}j^n, \ \ \ 1 \le j \le n-1.$$Since the square matrix on the LHS of the above is invertible (Vandermonde), we get that each $f^{(k)}(x), \ \ 1 \le k \le n-1,$ is a linear combination of $g_1(x), \cdots , g_{n-1}(x).$ So, in order to prove that $\lim_{x\to\infty}f^{(k)}(x)=0$ for all $1 \le k \le n-1,$ we only need to show that $\lim_{x\to\infty}g_j(x)=0$ for all $1 \le j \le n-1$ and that's easy to see. For each $j,$ we have $\lim_{x\to\infty}(f(x+j)-f(x))=0$ because we're given that $\lim_{x\to\infty}f(x)$ is a finite number. Also, $\lim_{x\to\infty}f^{(n)}(y_j)=0$ because we're given that $\lim_{x\to\infty}f^{(n)}(x)=0$ and $x < y_j < x+j.$ Thus $\lim_{x\to\infty}g_j(x)=0$ for all $j.$

very nice solution

Nice....

It's natural to use Taylor's expansion. $$f(a+t)=f(a)+f'(a)\frac{t}{1!}+\dots +f^{(n-1)}(a)\frac{t^{n-1}}{(n-1)!}+f^{(n)}(\xi)\frac{t^n}{n!}$$where $\xi\in (a,a+t).$ It means $$\frac{f'(a)}{1!}t+\dots +\frac{f^{(n-1)}(a)}{(n-1)!}t^{n-1} = f(a+t)-f(a)-f^{(n)}(\xi)\frac{t^n}{n!}\qquad (1)$$We take $a$ large enough and vary $t$ in the interval $[0,1].$ So, the RHS of $(1)$ can be made as small as we want as $a\to\infty$ and we want to show all $f^{(k)}(a), k=1,2,\dots,n-1$ are also small. The following lemma shows why it should hold. It has also a standalone importance. For example, the idea behind it was used in Putnam 1999, A5. Lemma. Let $r\in \mathbb{N}.$ There exists a constant $C$ (depending only on $r$) such that for any polynomial $P(t)=a_0t^r+a_1t^{r-1}+\dots a_{r-1}t+a_r$ we have $$|a_0|+|a_1|+\dots+|a_r|\le C\max_{t\in[0,1]} |P(t)|$$Proof. Consider the vector space $L_r$ of all polynomials of degree at most $r.$ It can be normed by any of the following two norms $$\|p\|_1:= |a_0|+|a_1|+\dots+|a_r|\,;\, \|p\|_2:=\max_{t\in[0,1]}|p(t)|$$for any $p\in L_r, p(t)=a_0t^r+a_1t^{r-1}+\dots a_{r-1}t+a_r$. Since $L_r$ is a finite dimensional vector space, any two norms are equivalent, which means there exist constants $C_1,C_2>0$ depending eventually only on $r$ so that $$C_1\|p\|_2\le \|p\|_1\le C_2\|p\|_2,\forall p\in L_r$$which completes the proof. $\blacksquare$ Back to the original problem. Taking into account $(1)$ and the given conditions, for any $\varepsilon>0$ we can take large enough $A>0$ such that for all $a>A$ and $t\in[0,1]$ it holds $$\left|\frac{f'(a)}{1!}t+ \frac{f''(a)}{2!}t^2 +\dots +\frac{f^{(n-1)}(a)}{(n-1)!}t^{n-1}\right|<\varepsilon$$By the above lemma we get $$\frac{\left|f'(a)\right|}{1!} +\frac{\left|f''(a)\right|}{2!}+\dots +\frac{\left|f^{(n-1)}(a)\right|}{(n-1)!}<C\varepsilon$$where $C$ is a constant, depending only on $n$. It implies $$\lim_{a\to\infty}f^{(k)}(a)=0, k=1,2,\dots,n-1.$$ Remark. Some other properties of polynomials, similar to that used in the above lemma.

I'm one of the contributors to this problem

I'm pretty sure this is an old problem. It was solved in a SEEMOUS training session in my university last year.