Show that $ \Phi(x,z) = x^{\delta} \cdot \prod_{p \le z} \bigg(1-\frac{1}{p^{\delta}}\bigg)^{-1}$

Question

I am stuck at the following exercise:

Let $\Phi(x, z)$ be the number of $n \le x$ all of whose prime factors are less than or equal to $z$. Prove that for any $\delta > 0$ holds

$$ \Phi(x,z) \le x^{\delta} \cdot \prod_{p \le z} \bigg(1-\frac{1}{p^{\delta}}\bigg)^{-1}.$$

I recognise the similarity of $\prod_{p \le z} \bigg(1-\frac{1}{p^{\delta}}\bigg)^{-1}$ to the Euler Product and if I am not mistaken it should thus hold:

$$\prod_{p \le z} \bigg(1-\frac{1}{p^{\delta}}\bigg)^{-1} = \prod_{p \le z} \frac{1}{1-p^{-\delta}}$$

But I do not see how this could help me.

Answer 1

$$\prod_{p \le z} \bigg(1-\frac{1}{p^{\delta}}\bigg)^{-1}=\sum_{n\ge 1,LargestPrimeFactor(n)\le z} n^{-\delta}$$ If $n\le x$ then $x^\delta n^{-\delta}$..