Let \(p\), \(q\), and \(r\) be distinct prime numbers.
WLOG, we may assume that \(p < q < r\).
Let \(G\) be a group with order \(pqr\).
We shall be appealing to the Sylow theorems without mentioning it explicitly.
If any of \(n_p\), \(n_q\), or \(n_r\) is equal to \(1\), then we know that \(G\) is not simple.
For sake of contradiction assume that each of the above is strictly greater than \(1\).
As \(n_r \mid pq\) and \(p, q < r\), we get that \(n_r = pq\). (Since we have assumed that \(n_r > 1\).)
Thus, there are \(pq\) Sylow-\(r\) subgroups of \(G\). Now, note that each such Sylow-\(r\) subgroup has order \(r\), a prime and thus, the intersection of two distinct Sylow-\(r\) subgroups must be trivial, id est, \((1)\). Thus, the number of elements having order \(r\) equals \(o_r = pq(r-1)\).
Now, \(n_q > 1\) and \(n_q \mid pr\). Thus, \(n_q \in \{p, r, pr\}\). However, \(n_q \equiv 1 \mod q\) and thus, \(n_q \neq p\). This gives us that \(n_q \ge r\). Thus, the number of elements having order \(q\) equals \(o_q \ge r(q-1)\).
Lastly, similar argument as earlier gives us that \(o_p \ge q(p-1)\).
Note that \(o_r\), \(o_q\), and \(o_p\) are counting distinct non-identity elements and thus,
\(|G| \ge o_r + o_q + o_p + 1 \ge pq(r-1) + r(q-1) + q(p-1)\)
\(\implies |G| \ge pqr + rq - r - q + 1 = pqr + \underbrace{(r-1)(q-1)}_{>0} > pqr\).
Thus, we have a contradiction as \(|G| = pqr\) and we are done!