Why are the integrands dominated by $\alpha f$ This is on page 32 of Rudin's Real and Complex Analysis, 3rd Edition:

Suppose $\mu$ is a positive measure on $X$, $f: X \rightarrow [0, \infty]$ is measurable, $\int_X f d\mu = c$, where $0<c<\infty$, and $\alpha$ is a constant. Prove that$$\lim_{n \rightarrow \infty} \int_X n \log[1+(f/n)^{\alpha}]d \mu = \begin{cases}
\infty & \text{ if } 0 < \alpha <1, \\ 
c & \text{ if } \alpha=1, \\ 
0 & \text{ if } 1 < \alpha < \infty.
\end{cases}$$

The hint says "if $\alpha \geq 1$, the integrands are dominated by $\alpha f$". But why?
Thanks a lot.

To be proven: $n\log(1+(t/n)^\alpha)\leqslant\alpha t$, for every $t\geqslant0$ and $n\gt0$, with $\alpha\geqslant1$.
Step 1 Replace $t$ by $nt$, hence it suffices to prove that $\log(1+t^\alpha)\leqslant\alpha t$, for every $t\geqslant0$.
Step 2 Show that, if $\alpha\geqslant1$, then $t^{\alpha-1}\leqslant1+t^\alpha$ for every $t\geqslant0$. (Hint: consider separately the cases $t\leqslant1$ and $t\geqslant1$.)
Step 3 Compute the derivative of the function $u:t\mapsto\log(1+t^\alpha)-\alpha t$.
Step 4 Compute $u(0)$ and conclude.
