Three of the previous exercise (here, here, and here, Section 10.9, Exercises #17 – #19) suggest a formula
![\[ \sum_{n=0}^{\infty} \binom{n+k}{k} x^n = \frac{1}{(1-x)^{k+1}}, \qquad \text{where} \ \ \binom{n+k}{k} = \frac{(n+1)(n+2) \cdots (n+k)}{k!}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-200879bdbea0dfced0a90afa0ca564e0_l3.png "Rendered by QuickLaTeX.com")
Use mathematical induction to prove this formula without justifying the formal manipulations with the series.
Proof. For the case 
we already established in the previous exercises
![\[ \sum_{n=0}^{\infty} (n+1)x^n = \frac{1}{1-x^2} \quad \implies \quad \sum_{n=0}^{\infty} \binom{n+k}{n} x^n = \frac{1}{(1-x)^{k+1}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3d5824d509839ea91b28b2cbdb8f87b5_l3.png "Rendered by QuickLaTeX.com")
Thus, the statement holds for the case . Assume then that it holds for some 
. Then, we have
![\[ \sum_{n=0}^{\infty} \binom{n+j}{j} x^n = \frac{1}{(1-x)^{j+1}}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-bd016ee6c028c5ea3bb9bc61c1bf7d8c_l3.png "Rendered by QuickLaTeX.com")
Differentiating both sides we then have
![\begin{align*} && \sum_{n=0}^{\infty} \binom{n+j}{j} nx^{n-1} &= \frac{j+1}{(1-x)^{j+2}} \\[9pt] \implies && \sum_{n=0}^{\infty} \frac{n(n+1) \cdots (n+j)}{(j+1)!} x^{n-1} &= \frac{1}{(1-x)^{j+2}} \\[9pt] \implies && \sum_{n=1}^{\infty} \frac{n(n+1) \cdots (n+j)}{(j+1)!} x^{n-1} &= \frac{1}{(1-x)^{j+2}} \\[9pt] \implies && \sum_{n=0}^{\infty} \frac{(n+1)(n+2) \cdots (n+j+1)}{(j+1)!} x^n &= \frac{1}{(1-x)^{j+2}} \\[9pt] \implies && \sum_{n=0}^{\infty} \binom{n+j+1}{j+1} x^n &= \frac{1}{(1-x)^{j+2}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f3e4b8e9f56083378de44a909ecd96a6_l3.png "Rendered by QuickLaTeX.com")
Thus, the statement holds for all positive integers 