Multiplying by $e^t$, this is equivalent to showing that $\displaystyle{0\leq 1-e^t\left(1-\frac{t}{n}\right)^n\leq \frac{t^2}{n}}$. Let $\displaystyle{f(t)=1-e^t\left(1-\frac{t}{n}\right)^n}$. Then \displaystyle{f'(t)=\frac{t}{n}e^t\left(1-\frac{t}{n}\right)^{n-1}\geq 0} for all $t\in[0,n]$, and $f(0)=0$, so $f(t)\geq 0$ for all $t\in[0,n]$, proving the first inequality.
For the second inequality, first suppose $n\geq 2$. Consider $g(t)=\frac{t^2}{n}$, and note that by the racetrack principle it suffices to show that f'(t)\leq g'(t) for all $t\in[0,n]$. This reduces to showing that $\displaystyle{e^t\left(1-\frac{t}{n}\right)^{n-1}\leq 2}$ for all $t\in[0,n]$. If we let $\displaystyle{h(t)=e^t\left(1-\frac{t}{n}\right)^{n-1}}$, then \displaystyle{h'(t)=e^t\left(1-\frac{t}{n}\right)^{n-2}\cdot\frac{1-t}{n}}, which shows that $h$ reaches its maximum at $t=1$. Thus the second inequality, in the case $n\geq 2$, is reduced to proving that $\displaystyle{e\left(1-\frac{1}{n}\right)^{n-1}\leq 2}$. You can check that the inequality holds when $n=2$, and you can show that $\displaystyle{\left(1-\frac{1}{n}\right)^{n-1}}$ decreases as $n$ increases.
This just leaves the $n=1$ case for the second inequality, or $1-e^t(1-t)\leq t^2$. Taking the derivative of $t^2-1+e^t(1-t)$ shows that this function increases on $(0,\ln 2)$ and decreases on $(\ln(2),1)$, and it is $0$ when $t=0$ or $t=1$. This implies that $t^2-1+e^t(1-t)\geq 0$ for all $t\in[0,1]$. Alternatively, when $t<1$, factoring out $1-t$ reduces the inequality to the well known fact that $e^t\geq 1+t$.