Help in proving that $\bigl (n^\alpha x^n (1-x) \bigr)_{n=0}^{\infty}$ is convergent. I need some help in the following question:

Let $\ 0&lt;\alpha \in \mathbb R$. If $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ a  sequence of functions as $\forall n\in\mathbb N \ \ \ f_n:[0,1]\to \mathbb R$:$$\forall n\in\mathbb N, \ \ \forall x\in\mathbb [0,1]: \ \ \ f_n(x)=n^\alpha x^n (1-x)$$

Q:  1. Prove that $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ is pointwise convergent in $[0,1]$.

Prove that $\bigl (f_n(x) \bigr)_{n=0}^{\infty}$ is uniformly convergent iff $\alpha&lt;1$.

My thought till now: For the first part I tried to prove $\lim_{n\to\infty}n^\alpha x^n (1-x)=0$ for a constant $x$ but the $n^\alpha$ is confusing me and I don't know how to continue.
Thanks in advance
Hint:
Just show the logarithm of
$$\ln(n^\alpha x^n)=\alpha\ln n+n\ln x \to -\infty.$$
We can write estimation:
$$|f_n(x)|=|n^\alpha x^n (1-x)| \leqslant n^\alpha$$
from which for $\alpha &lt; 0$ you have as 1 so 2.
For case $0 &lt; \alpha &lt; 1$ It's easy find, that we have maximum in  $x=\frac{n}{n+1}$ and then  use criterion
$$\lim_{n \to \infty} \sup_{}|n^\alpha x^n (1-x)|=\lim_{n \to \infty} n^{\alpha} \left( \frac{n}{n+1}\right)^n \cdot \frac{1}{n+1}=0$$