a. Induct on the degree $n$ of the polynomial $P(x) = \sum_{k=0}^{n} a_k x^k$. It is immediate if $n=0$. Since the degree of $(x^k)$, $k < n$, is $k$, the coefficient on $(x^n)$ on $P(x)$ is $a_n \cdot n!$. Now consider the polynomial $P(x) - a_n \cdot n! (x^n)$. This is a polynomial with degree less than $n$, so $P(x) - a_n \cdot n! (x^n)$ can be uniquely represented as a linear combination of the polynomials $(x^k)$, and the result follows.

b. We have $x^n = \sum_{k=0}^{n} c(n, k) \binom{x}{k}$, so
$$\begin{align*} x^{n+1} &= \sum_{k=0}^{n} c(n,k) \binom{x}{k} x \ &= \sum_{k=0}^{n} c(n,k) \binom{x}{k} (x+1) - \sum_{k=0}^{n} c(n,k) \binom{x}{k} \ &= \sum_{k=0}^{n} c(n,k) (k+1) \binom{x+1}{k+1} - \sum_{k=0}^{n} c(n,k) \binom{x}{k} \ &= \sum_{k=0}^{n} c(n,k) (k+1) \left( \binom{x}{k+1} + \binom{x}{k} \right) - \sum_{k=0}^{n} c(n,k) \binom{x}{k} \ &= \sum_{k=0}^{n} (c(n,k-1)k + (k+1-1)c(n,k)) \binom{x}{k} \ &= \sum_{k=0}^{n} k(c(n,k-1) + c(n,k)) \binom{x}{k} \end{align*}$$
in which we consider $c(n, -1) = 0$.
Thus $c(n+1, k) = k(c(n, k-1) + c(n, k)) \iff \frac{c(n,k)+c(n,k-1)}{c(n+1,k)} = \frac{1}{k}$, and
$\frac{c(n,k+1)+c(n,k)}{c(n+1,k+1)} = \frac{1}{k+1}$.