Question 352111
f(x)=f(a)+f'(a)(x-a)+f''(a)(x-a)^2+f'''(a)(x-a)^3+...+f^(10)(a)(x-a)^10
In this case 
f(x)=f'(x)=f''(x)=...={{{f^(10)(x)=e^x}}}
{{{a=0}}}
Then just substitute,
{{{f(x)=e^(a)+e^(a)(x-a)+e^(a)(x-a)^2+e^(a)(x-a)^3}}}+...+{{{e^(a)(x-a)^10}}}
{{{f(x)=sum(x^i,i=0,10)}}}