$f_n^\alpha(x) = n^\alpha x^n$ converges almost everywhere
Solution 1:
Rewrite it as (for $x\in(0,1)$)
$$
f_n^\alpha(x) = e^{\alpha \log(n)}\cdot e^{n\log(x)} = e^{\alpha \log(n) + n\log(x)}
$$
Now for $x\in(0,1)$, $\log(x) = -\varepsilon$ for some $\varepsilon >0$.
And the limit
$$
\lim_{n\to\infty} \big(\alpha\log(n) -\varepsilon n\big) = -\infty
$$
So $f_n^\alpha(x) \to 0$ for $x\in(0,1)$.
Now when $x=0$ you get that
$$f_n^\alpha(0) = n^\alpha \cdot 0 = 0$$
Lastly when $x=1$ you get that $$f_n^\alpha(1) = n^\alpha\to\infty$$ meaning it diverges only when $x=1$.
Hence $f_n^\alpha$ converge almost everywhere.