To show: $$\forall n\in\mathbb{N}_0,\forall x\in\mathbb{R}:\sum_{k=0}^{n}(-1)^k\binom{x}{k}=(-1)^n\binom{x-1}{n}$$ $\underline{Basecase}$
$n=0$ $$\sum_{k=0}^{0}(-1)^k\binom{x}{k}=(-1)^0\binom{x}{0}=1=(-1)^0\binom{x-1}{0}$$
$\underline{Assumption}$
For some $n\in\mathbb{N}_0$ and a fixed but arbitary $x\in\mathbb{R}$ the following holds:$$\sum_{k=0}^{n}(-1)^k\binom{x}{k}=(-1)^n\binom{x-1}{n}$$ $$\iff\sum_{k=0}^{n}(-1)^k\prod_{i=1}^{k}\frac{x+1-i}{i}=(-1)^{n}\prod_{i=1}^{n}\frac{x-i}{i}$$ $$\iff\sum_{k=0}^{n}\prod_{i=1}^{k}\frac{-x-1+i}{i}=\prod_{i=1}^{n}\frac{-x+i}{i}$$ $$\iff\sum_{k=0}^{n}\frac{1}{k!}\prod_{i=1}^{k}(-x-1+i)=\frac{1}{n!}\prod_{i=1}^{n}(-x+i)$$ $\underline{Claim}$
For $(n+1)\in\mathbb{N}_0$ and a fixed but arbitary $x\in\mathbb{R}$ the following holds:$$\sum_{k=0}^{n+1}\frac{1}{k!}\prod_{i=1}^{k}(-x-1+i)=\frac{1}{(n+1)!}\prod_{i=1}^{n+1}(-x+i)$$ $\underline{Step}$
Let $x\in\mathbb{R}$ be arbitary but fixed. Let $(n+1)\in\mathbb{N}_0$. We have to show that the equality$$\sum_{k=0}^{n+1}\frac{1}{k!}\prod_{i=1}^{k}(-x-1+i)=\frac{1}{(n+1)!}\prod_{i=1}^{n+1}(-x+i)$$ holds. $$\iff\frac{1}{(n+1)!}\prod_{i=1}^{n+1}(-x-1+i)+\sum_{k=0}^{n}\frac{1}{k!}\prod_{i=1}^{k}(-x-1+i)=\frac{1}{(n+1)!}\prod_{i=1}^{n+1}(-x+i)$$ Now use our assumption to get $$\iff\frac{1}{(n+1)!}\prod_{i=1}^{n+1}(-x-1+i)+\frac{1}{n!}\prod_{i=1}^{n}(-x+i)=\frac{1}{(n+1)!}\prod_{i=1}^{n+1}(-x+i)$$ $$\iff\frac{(-x+n)}{(n+1)!}\prod_{i=1}^{n}(-x-1+i)+\frac{1}{n!}\prod_{i=1}^{n}(-x+i)=\frac{(-x+n+1)}{(n+1)!}\prod_{i=1}^{n}(-x+i)$$ $$\iff\frac{(-x+n)}{(n+1)}\prod_{i=1}^{n}(-x-1+i)+\prod_{i=1}^{n}(-x+i)=\frac{(-x+n+1)}{(n+1)}\prod_{i=1}^{n}(-x+i)$$ Im kinda lost at the end. Could you say if this is going into the right direction?
Since both sides are polynomials of degree $n$ in $x$, it suffices to show that the equality holds for $x=1,2, ... , n+1$. For that, use the fact that the alternating sum of binomial coefficients is 0.