Answer
$f(x)=x^{4}+x^{3}+x^{2}+x+1$
$a_{n}=1$
Work Step by Step
Let $f(x)=ax^{4}+bx^{3}+cx^{2}+dx+e$
$f'(x)=4ax^{3}+3bx^{2} + 2cx+d$
$f''(x)=12ax^{2}+6bx+2c$
$f'''(x)=24ax+6b$
$f^{4}(x)=24a$
The desired polynomial hence by the hypothesis:
$f(0)=1 \rightarrow e=1$
$f'(0)=1 \rightarrow d=1$
$f''(0)=2 \rightarrow c=1$
$f'''(0)=6 \rightarrow b=1$
$f^{4}(0)=24 \rightarrow a=1$
The polynomial of degree 4 is
$f(x)=x^{4}+x^{3}+x^{2}+x+1$
$f(x)=x^{n}+x^{n-1}+x^{n-2}+...+x+1$
$f^{(n)}(0)=n!$
$n!a_{n}=n!$
$a_{n}=1$
