For pointwise convergence, you fix $x\in [0,1] $ and you compute $\lim_ {n\to+\infty}f(x) $.
There will be pointwise convergence in the set containing $x $ for which the limit exists $(\in\mathbb R) $.
in your example,
for $x=0$, the limit is zero.
for $x=1$, it is zero.
for $0 <x <1$, write
$(1-x)^n=e^{n\ln (1-x)} $ and you will find zero since exponential is faster than polynomial.
thus all the limits are zero in $[0,1] $ .
$(f_n) $ converges (pointwise) to function $0$ at $[0,1] $
For uniform convergence
find the maximum of $|f_n (x)-0|=f_n (x)$ at $[0,1] $.
$$f'_n (x)=n^2 (1-x)^{n-1}(1-x-nx) $$
it is attained at $$x_n=\frac {1}{n+1} $$
$$f (x_n)=n (\frac {n}{n+1})^{n+1}$$
$$\lim_{n\to+\infty}f_n (x_n)=+\infty$$
the convergence is not uniform at $[0,1] $.
For bounded functions $f:S\to \Bbb R$ and $f_n:S\to \Bbb R$ let $$\|f-f_n\|=\sup_{x\in S}|f(x)-f_n(x)|.$$ Consider $\|f-f_n\|$ to be a measurement of how closely $f_n$ approximates $f,$ not at any one $x\in S,$ but as an over-all measurement over all of $S.$ Consider $f$ and $f_n$ as single objects, and $\|f-f_n\|$ as the distance from $f$ to $f_n.$
The sequence $(f_n)_n$ converges uniformly iff $\|f-f_n\|\to 0$ as $n\to \infty.$
If $(f_n)_n$ converges uniformly to $f$ then for any $y\in S$ we have $$|f(y)-f_n(x)|\leq \|f-f_n\|,$$ so $f_n(y)\to f(y)$ as $n\to \infty.$
If $f_n(y)\to f(y)$ for each $y\in S$ we say that $f_n$ converges point-wise to $f.$ So uniform convergence implies point-wise convergence (but not vice-versa). If we are testing whether $(f_n)_n$ converges uniformly to some (any) $f$, we can, as a first step, identify the "candidate" $f$ by evaluating $\lim_{n\to \infty}f_n(y)$ for each $y\in S.$
With $S=[0,1)$ and $f_n(x)=nx^n,$ then $f_n(y)\to 0$ for each $y\in S.$ By the previous paragraph, if $(f_n)_n$ did converge uniformly to a function $f,$ then we would have $f(y)=\lim_{n\to \infty}f_n(x)=0$ for each $y\in S$..... But for any $n\in \Bbb N$ we can find some $y_n \in S$ with $y_n$ close enough to $1$ that $(y_n)^n>1/2.$ Then $f(y_n)>n/2$ and $f(y)=0$ so $$\|f-f_n\|\geq |f(y_n)-f_n(y_n)|=|0-f_n(y_n)|>n/2.$$ So the sequence $(f_n)_n$ in the Q is not a uniformly convergent sequence..
BTW . An important result that holds for all metric spaces $S, T$ (not just sub-spaces of $\Bbb R$ ), is that if $(f_n:S\to T)_{n\in \Bbb N}$ is a sequence of continuous functions converging uniformly to $f:S\to T$, then $f$ is continuous.
Best Answer
In fact, since $(\forall n\in\mathbb N):\frac1n\in(0,1)$, you can use the same argument with $\frac1n$. Since$$(\forall n\in\mathbb N):f\left(\frac1n,n\right)=\frac12,$$$(f_n)_{n\in\mathbb N}$ does not converge uniformly to the null function and therefore, since it converges pointwise to that function, it doesn't converge uniformly at all.