Answer
$y = {{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$
Work Step by Step
Let $y = P\left( x \right) = \mathop \sum \limits_{n = 0}^\infty {a_n}{x^n}$.
$y = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \cdot\cdot\cdot$
With $P\left( 0 \right) = 1$, we get ${a_0} = 1$.
The derivatives are
$y' = \mathop \sum \limits_{n = 1}^\infty n{a_n}{x^{n - 1}} = {a_1} + 2{a_2}x + 3{a_3}{x^2} + \cdot\cdot\cdot$
$y{\rm{''}} = \mathop \sum \limits_{n = 2}^\infty n\left( {n - 1} \right){a_n}{x^{n - 2}} = 2{a_2} + 3\cdot2{a_3}x + \cdot\cdot\cdot$
Shifting the indices, we obtain
$y' = \mathop \sum \limits_{n = 0}^\infty \left( {n + 1} \right){a_{n + 1}}{x^n} = {a_1} + 2{a_2}x + 3{a_3}{x^2} + \cdot\cdot\cdot$
$y{\rm{''}} = \mathop \sum \limits_{n = 0}^\infty \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^n} = 2{a_2} + 3\cdot2{a_3}x + \cdot\cdot\cdot$
Substituting $y'$ and $y{\rm{''}}$ in the Laguerre differential equation:
$xy{\rm{''}} + \left( {1 - x} \right)y' - y = 0$
gives
$x\mathop \sum \limits_{n = 0}^\infty \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^n} + \left( {1 - x} \right)\mathop \sum \limits_{n = 0}^\infty \left( {n + 1} \right){a_{n + 1}}{x^n} - \mathop \sum \limits_{n = 0}^\infty {a_n}{x^n} = 0$
$\mathop \sum \limits_{n = 0}^\infty \left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}}{x^{n + 1}} + \mathop \sum \limits_{n = 0}^\infty \left( {n + 1} \right){a_{n + 1}}{x^n} - \mathop \sum \limits_{n = 0}^\infty \left( {n + 1} \right){a_{n + 1}}{x^{n + 1}} - \mathop \sum \limits_{n = 0}^\infty {a_n}{x^n} = 0$
$\mathop \sum \limits_{n = 0}^\infty \left[ {\left( {n + 2} \right)\left( {n + 1} \right){a_{n + 2}} - \left( {n + 1} \right){a_{n + 1}}} \right]{x^{n + 1}} + \mathop \sum \limits_{n = 0}^\infty \left[ {\left( {n + 1} \right){a_{n + 1}} - {a_n}} \right]{x^n} = 0$
Shifting the index gives
$\mathop \sum \limits_{n = 1}^\infty \left[ {\left( {n + 1} \right)\left( n \right){a_{n + 1}} - n{a_n}} \right]{x^n} + \mathop \sum \limits_{n = 0}^\infty \left[ {\left( {n + 1} \right){a_{n + 1}} - {a_n}} \right]{x^n} = 0$
The differential equation is satisfied if
$\mathop \sum \limits_{n = 1}^\infty \left[ {n\left( {n + 1} \right){a_{n + 1}} - n{a_n}} \right]{x^n} = - \mathop \sum \limits_{n = 0}^\infty \left[ {\left( {n + 1} \right){a_{n + 1}} - {a_n}} \right]{x^n}$
We write out the first few terms on each side of this equation:
$\left( {2{a_2} - {a_1}} \right)x + \left( {6{a_3} - 2{a_2}} \right){x^2} + \left( {12{a_4} - 3{a_3}} \right){x^3} + \left( {20{a_5} - 4{a_4}} \right){x^4} + \left( {30{a_6} - 5{a_5}} \right){x^5}$
$ = - \left( {{a_1} - {a_0}} \right) - \left( {2{a_2} - {a_1}} \right)x - \left( {3{a_3} - {a_2}} \right){x^2} - \left( {4{a_4} - {a_3}} \right){x^3} - \left( {5{a_5} - {a_4}} \right){x^4} - \left( {6{a_6} - {a_5}} \right){x^5}$
Matching up the coefficients of ${x^n}$, we obtain
${a_1} - {a_0} = 0$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${a_1} = {a_0}$
$2{a_2} - {a_1} = - 2{a_2} + {a_1}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ $4{a_2} = 2{a_1}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${a_2} = \dfrac{1}{2}{a_1}$
$6{a_3} - 2{a_2} = - 3{a_3} + {a_2}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ $9{a_3} = 3{a_2}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${a_3} = \dfrac{1}{3}{a_2}$
$12{a_4} - 3{a_3} = - 4{a_4} + {a_3}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ $16{a_4} = 4{a_3}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${a_4} = \dfrac{1}{4}{a_3}$
$20{a_5} - 4{a_4} = - 5{a_5} + {a_4}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ $25{a_5} = 5{a_4}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${a_5} = \dfrac{1}{5}{a_4}$
$30{a_6} - 5{a_5} = - 6{a_6} + {a_5}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ $36{a_6} = 6{a_5}$ ${\ \ \ }$ $ \Rightarrow $ ${\ \ \ }$ ${a_6} = \dfrac{1}{6}{a_5}$
Since ${a_0} = 1$, so we obtain
$\begin{array}{*{20}{c}}
n&{}&{{a_n}}\\
1&{}&1\\
2&{}&{\dfrac{1}{2}}\\
3&{}&{\dfrac{1}{3}\cdot\dfrac{1}{2}}\\
4&{}&{\dfrac{1}{4}\cdot\dfrac{1}{3}\cdot\dfrac{1}{2}}\\
5&{}&{\dfrac{1}{5}\cdot\dfrac{1}{4}\cdot\dfrac{1}{3}\cdot\dfrac{1}{2}}\\
6&{}&{\dfrac{1}{6}\cdot\dfrac{1}{5}\cdot\dfrac{1}{4}\cdot\dfrac{1}{3}\cdot\dfrac{1}{2}}
\end{array}$
In general: ${a_n} = \dfrac{1}{{n!}}$.
Substituting ${a_n} = \dfrac{1}{{n!}}$ in $y = P\left( x \right) = \mathop \sum \limits_{n = 0}^\infty {a_n}{x^n}$ gives
$y = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$
From Table 2 of Section 11.7, we know that
${{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$
Therefore, the solution is
$y = {{\rm{e}}^x} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$
