Taylor series for $\frac{1}{(1-x)^{n}}$

Question

So I am trying to find the dimension of the subspace of homogeneous polynomials of degree m $P_m$ of $n$ variables. For $\alpha = (\alpha_1, \dots, \alpha_n$), with $\alpha_1 +\cdots + \alpha_n = m.$
I have shown that $(1-t)^{-n} = \sum_{m=0}^{\infty} \delta_m t^m$ with $\delta_m = \binom{m+n-1}{n-1}$. But i just cant figure out why $\delta_m$ is the dimension. I would like to prove that $\delta_m = \# \left\{ \alpha \in \mathbb{N^n} | \alpha_1 + \cdots + \alpha_n = m \right\}$. So what I can figure out the next step would be something like \begin{align} (1-t)^{-n} &= (1+t+t^2+\cdots + t^m+ \cdots )^n \\ &= \sum_{\alpha \in \mathbb{N}^n} t^{\alpha_1}\cdots t^{\alpha_n} \\ &= \sum_{m=0}^{\infty} \# \left\{ \alpha \in \mathbb{N^n} | \alpha_1 + \cdots + \alpha_n = m \right\} t^m \end{align} and then I would be done. So I understand the first equality, but the second and third I am not sure where they come from?

Answer 1

Hint: You can easily prove, with an induction for nursery schools, that $$\biggl(\frac1{1-x}\biggr)^{\!(n)}=\frac{n!}{(1-x)^{n+1}},$$ so $$\frac 1{(1-x)^n}=\frac 1{(n-1)!}\,\biggl(\frac1{1-x}\biggr)^{\!(n-1)}.$$