For your first question, we're just factoring out an n! from each term. You can double check this by distributing: $n!\cdot (n+1) = n!\cdot n + n!\cdot 1 = n\cdot n! + n!$.

The second question can be a little more confusing. For reference, remember $\displaystyle \binom{A}{B} = \frac{A!}{B!\cdot (A-B)!}$. (I changed the variables on purpose just to make them stand our more compared to n's and k's.)

Now let's look at $\dfrac{(n+1)\cdot n!}{(k+1)!\cdot (n-k)!}$. What happens if $A = (n+1)$ and $B = (k+1)$? First, note $(n+1)\cdot n! = (n+1)!$ so the numerator of the fraction is just $A!$. For the denominator, $(k+1)! = B!$ and note $A-B = (n+1) - (k+1) = (n-k)$ so the other part of the denominator is $(A-B)!$.

Hence, the fraction $\displaystyle \frac{(n+1)\cdot n!}{(k+1)!\cdot (n-k)!} = \frac{A!}{B!\cdot (A-B)!} = \binom{A}{B}$.

Hope this helps!