Proof of $\log(n!) < (n-1)\log(n)$ 
Prove that $\log(n!) \leq (n-1)\log(n)$ directly and by induction.

I'm having a hard time with this one. I tried:
$$\log(n!) + \log(n+1) \leq (n-1)\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n) - \log(n) + \log(n+1) \leq n\log(n+1) - \log(n) + \log(n+1)$$
$$\log((n+1)!) \leq n\log(n+1) - \log(n) + \log(n+1) \leq n\log(n+1) + \log(n+1)$$
$$\log((n+1)!) \leq (n+1)\log(n+1)$$
But that's not what is asked, and I can't find a way of getting there.
 A: Hint: You can use $n!\leqslant n^{n-1}$
A: By induction you have,
$$\log(n!)+\log(n+1)\le (n-1)\log(n)+\log(n+1)\to$$
$$\log[(n+1)!]\le (n-1)\log(n)+\log(n+1)$$
Remmember that $\log()$ is a strictly increasing function, so $\log(n)< \log(n+1)$. It means,
$$(n-1)\log(n)\le (n-1)\log(n+1)\to$$
$$(n-1)\log(n)+\log(n+1)\le (n-1)\log(n+1)+\log(n+1)=n\log(n+1).$$
A: \begin{align}
\log(n!) &\leq (n-1)\log(n) \\
\iff \log(n) + \log(n!) &\leq \log(n^n) \\
\iff \log(nn!) &\leq \log(n^n)
\end{align}
Since $\log$ is an increasing function, and $nn!\leq n^n$, the third inequality holds true for all $n \in \mathbb{N}$. I would try proving that $nn! \leq n^n$ myself, and then you can compare with my solution:
\begin{align}
nn! &\leq n^n \iff n! \leq n^{n-1} \\[6pt]
n! &= \underbrace{n \cdot (n-1) \cdot (n-2) \cdot (n-3) \cdot \ldots \cdot 2}_{n-1 \text{ terms}} \\[6pt]
n^{n-1}&=\underbrace{n \cdot n \cdot n \cdot n \cdot \ldots \cdot n}_{n-1 \text{ terms}}
\end{align}
The case $n=1$ is a mild annoyance, but can be dealt with easily.
A: $\log(n!)$
$=\sum_{m=1}^{n}\log(m)$
$=\log(1)+\sum_{m=2}^{n}\log(m)$
$=\sum_{m=2}^{n}\log(m)$
$\log(m)$ is an increasing function
So
$\sum_{m=2}^{n}\log(m)$
$\le \sum_{m=2}^{n}\log(n)$
$ = (n-2+1)\log(n)$
$ = (n-1)\log(n)$
For induction proof:
Base case: $n=1$
$\log(1!) = 0$.
$(1-1)\log(1) = 0$.
So $\log(n!) \le (n-1)\log(n)$ is true for $n=1$.
Assume true for $n=k$
$\log(k!) \le (k-1)\log(k)$
Now prove true for $n=k+1$
$\log((k+1)!)$
$=\log(k+1) + \log(k!)$
$\le \log(k+1) + (k-1)\log(k)$ (by inductive assumption)
$\le \log(k+1) + (k-1)\log(k+1)$ (again since log is an increasing function)
$= ((k+1)-1)\log(k+1)$
So it's true for $n=k+1$
