Am I correct when I do this in questions about induction? I have a problem which is:
If $n$ is a fixed positive integer and greater than 1, show by induction that for each positive integer $r$, where $2\le r\le n$,
$$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{r-1}{n})$$

Denote this inequality(proposition) with variable r by $(\theta)_r$ 
I proved that when $r=2$, the above inequality is true, since
$$1-\frac{2(1-1)}{n}=1-\frac{2}{n} \le 1-\frac{1}{n}$$
Then I suppose for some $2\le k \le n-1$, $(\theta)_k$ is true, i.e.
$$1-\frac{k(k-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k-1}{n})$$Then I multiply both sides by $(1-\frac{k}{n})$, and I arrived at
$$1-\frac{k^2}{n}+\frac{k^2(k-1)}{n^2} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$$
After that I was stuck and I changed my proposition $(\theta)_r$ as follows:
$$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{r-1}{n})$$ and$$1-\frac{r(r+1)}{n} \le 1-\frac{r^2}{n}+\frac{r^2(r-1)}{n^2}$$
Then new proposition $(\theta_0)_r$ with variable $r$ holds when $r=2$, since
$$1-\frac{2(2+1)}{n}-1+\frac{2^2}{n}-\frac{2^2(2-1)}{n^2}=\frac{-4-2n}{n^2} \le 0$$
Then I suppose for some $2\le k \le n-1$, $(\theta_0)_k$ is true and go back to
$$1-\frac{k^2}{n}+\frac{k^2(k-1)}{n^2} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$$  $$ \Rightarrow1-\frac{k(k+1)}{n}\le 1-\frac{k^2}{n}+\frac{k^2(k-1)}{n^2} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$$
And I conclude, by induction, that $$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{r-1}{n})$$ holds for all positive integer $r$, where $2\le r \le n$

Am I correct when I modify the proposition in this way?
I feel that it is true and I can't see any logical fallacies so far...
Please correct me if this is wrong

Maybe you could prove the inequality,
$$1-\frac{r(r+1)}{n} \le 1-\frac{r^2}{n}+\frac{r^2(r-1)}{n^2}$$
directly (without induction). Infact by expanding you obtain
$$1-\frac{r^2}{n}-\frac{r}{n}\le 1-\frac{r^2}{n}+\frac{r^3}{n^2}-\frac{r^2}{n^2},$$
that is, $-1\le \frac{r^2}{n}-\frac{r}{n},$ or $-n\leq r(r-1)$ which trivially holds because $2\le r\le n$.
