(Somewhat) generalised mean value theorem

Hint: You might write the difference as $$ \int_0^1 (1-t)^n (f(t) - f(t_0)) dt $$ and ask what happens if $f(t_0)$ is the minimum or maximum of $f$.

Generally you have the following integral mean value theorem.

Theorem

If $f$ and $g$ are integrable functions with $f$ continuous and $g$ not changing sign, then hen there is some $c \in [a,b]$ such that $$ \int_a^b f(x) g(x) dx = f(c) \int_a^b g(x) dx.$$

Once you know the statement, it's quite straightfoward to prove. If you let $\gamma = \int_a^b g(x) dx$ and let $m,M$ be the min,max that $f$ achieves on $[a,b]$ (respectively), then $$ m\gamma \leq \int_a^b f(x) g(x) dx \leq M\gamma,$$ or rather $$ m \leq \frac{\int_a^b f(x) g(x) dx}{\gamma} \leq M.$$ By the intermediate value theorem, $f$ takes every value from $m$ to $M$ within $[a,b]$, and so there is some $c \in [a,b]$ such that $$ f(c) = \frac{\int_a^b f(x) g(x) dx}{\gamma}.$$

Rearranging gives the theorem [except when $\gamma = 0$ --- but that's a straightforward exercise that I leave aside]. $\diamondsuit$

This applies to your problem by taking $f = f$ and $g(x) = (1-x)^n$ in the theorem above. Then in particular $$ \int_0^1 g(x) dx = \int_0^1 (1-x)^n dx = \frac{1}{n+1}.$$

This concludes the proof. $\spadesuit$