Power series solution to Laguerre differential equation

The question is asking to find some power series $P(x)$ that satisfies the Laguerre differential equation:

$$xy''+(1-x)y'-y=0\text,$$

with initial condition $P(0) = 1$.

I am aware the solution to the differential equation is $P(x) = \exp(x)$ via WolframAlpha, so I thus know the power series. However, I must find a proof of this. The problem must be solvable without using DE solving techniques, since those have not been covered. What I have tried is plugging a general power series into the DE and simplifying by taking derivatives of it, but that did not seem to help.

I would prefer a hint over a full solution.

Thank you.

Solution 1:

\begin{align} P(x)&=\sum_{n=0}^{\infty}a_nx^n\\ xP''(x)&=\sum_{n=1}^{\infty}(n-1)na_nx^{n-1} =\sum_{n=0}^{\infty}n(n+1)a_{n+1}x^{n}\\ (1-x)P'(x)&=\sum_{n=1}^{\infty}na_nx^{n-1}-\sum_{n=0}^{\infty}na_nx^{n} =\sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n}-\sum_{n=0}^{\infty}na_nx^{n}\\ a_n&=n(n+1)a_{n+1}+(n+1)a_{n+1}-na_n\quad(n\ge0)\\ a_{n+1}&=\frac{a_n}{n+1}\quad\quad(n\ge0)\\ P(0)&=1\rightarrow a_0=1\\ \therefore a_n&=\frac1{n!}\\ \end{align} , which also means that $P(x)=e^x$.