Why does the inequality $1-x^5 \leq 5(1-x)$ hold? If $x \in [0,1]$. It is obviously true, by analyzing the plot of these two functions. But how can one prove it analytically?
For $x\in[0,1]$ we have $0\leq x^n\leq 1$, $n\in\mathbb N$. Then we simply get that

$$
\frac{1-x^5}{1-x}=\frac{(1-x)(1+x+x^2+x^3+x^4)}{1-x}=1+x+x^2+x^3+x^4\leq 5.
$$

Multiplying by $1-x$ we show that the given relation holds true.
Replacing $x$ by $1+y$ gives
$$ (1+y)^5 - 5y - 1 = y^5 + 5y^4 + 10y^3 + 10y^2 &gt; 0 $$
when $y&gt;0$, i.e. when $x&gt;-1$, so $x^5 + 5(1-x) - 1 &gt; 0$, equivalent to your inequality. The same works if $5$ is replaced any positive integer.