$ p$ is a prime. First we have to prove that $ p^k$ does not divide $ k!$.To show that put $ r$ so that $ p^r =< k < p^{r+1}$. the highest power of $ p$ dividing$ k!$ is $ p^r$(p adic valuation), such that $ r = [k/p] + [k/p^2] + [k/p^3] + ... + [k/p^r] ≤ k/p + k/p^2 + ... + k/p^r = k(1 - 1/p^r)/(p - 1) < k$. Now suppose there is a rational root $ x$ of the polynomial. if we take $ x = a/b$, where$ a$and$ b$ are coprime (a^b=1) integers. so $ a^n + na^{n-1}b + ... + (n-1)! a b^{n-1} + n! b^n = 0$. So an must be divisible by$ b$. But a and b are coprime, so$ b$ must be $ 1$, in other words, the root must be an integer. take $ p$ a prime dividing $ n$. Then p must divide $ a^n$ and thus $ a$. But we have$ - n! = n! (a^n/n! + a^{n-1}/(n-1)! + ... + a^1/1! )$. Each term inside the parentheses is divisible by$ p$, because $ p^k$ divides $ a^k$ but not $ k!$ So a higher power of $ p$ divides the $ rhs$ than the$ lhs$. Contradiction!

Notice that: $\sum_{k=0}^{\infty}\frac{x^k}{k!}=1+x+...=e^x$ $\Longleftrightarrow$ $e^x-1=\sum_{k=1}^{\infty}\frac{x^k}{k!}$ Furthermore lets take: $\lim_{n\to\infty}\sum^n_{k=1} \frac{x^k}{k!}=\sum_{k=1}^{\infty}\frac{x^k}{k!}$ It follows that $\sum_{k=1}^{\infty}\frac{x^k}{k!}+1=0$ but we know that $\sum_{k=1}^{\infty}\frac{x^k}{k!}=e^x-1$ $e^x-1+1=0$ $e^x=0$ There are no such real numbers $x$ such that this statement is true which implies that there are no real roots which implies that there are no rational roots. And we are done!