Answer
This shows the first three terms of the Maclaurin series:
${{\rm{e}}^{{x^{20}}}} = 1 + {x^{20}} + \dfrac{{{x^{40}}}}{{2!}} + \cdot \cdot \cdot $
We show that the coefficients of the series are zero for $1 \le k \le 19$. Therefore, we conclude that ${f^{\left( k \right)}}\left( 0 \right) = 0$ for $1 \le k \le 19$.
Work Step by Step
From Table 2, we get
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} = 1 + x + \dfrac{{{x^2}}}{{2!}} + \dfrac{{{x^3}}}{{3!}} + \dfrac{{{x^4}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
Substituting ${x^{20}}$ for $x$ in the series above gives
${{\rm{e}}^{{x^{20}}}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^{20n}}}}{{n!}} = 1 + {x^{20}} + \dfrac{{{x^{40}}}}{{2!}} + \dfrac{{{x^{60}}}}{{3!}} + \dfrac{{{x^{80}}}}{{4!}} + \cdot\cdot\cdot$, ${\ \ \ }$ converges for all $x$.
According to Maclaurin series: $f\left( x \right) = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}$.
For the first three terms, we get
$\dfrac{{{f^{\left( 0 \right)}}\left( 0 \right)}}{{0!}} = 1$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${f^{\left( 0 \right)}}\left( 0 \right) = 1$
$\dfrac{{{f^{\left( {20} \right)}}\left( 0 \right)}}{{20!}} = 1$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${f^{\left( {20} \right)}}\left( 0 \right) = 20!$
$\dfrac{{{f^{\left( {40} \right)}}\left( 0 \right)}}{{40!}} = \dfrac{1}{{2!}}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${f^{\left( {40} \right)}}\left( 0 \right) = \dfrac{{40!}}{{2!}}$
Notice that the coefficients of the series are zero for $1 \le k \le 19$. Therefore, we conclude that ${f^{\left( k \right)}}\left( 0 \right) = 0$ for $1 \le k \le 19$.
