Posting a second answer because there is a completely different, more direct proof.
Note that subtracting the $(n+2)^n$ on the right over, this is equivalent to proving
$$
\sum_{i=0}^{n+1} (-1)^{n-i}\binom{n+1}i(i+1)^n\stackrel{?}=0
$$
which after multiplying both sides by $(-1)^n$ equals
$$
\sum_{i=0}^{n+1} (-1)^{i}\binom{n+1}i(i+1)^n \stackrel{?}=0\tag{1}
$$
To prove this, consider the polynomial
$$
\sum_{i=0}^{n+1} \binom{n+1}i(i+1)^nx^i \tag{2}
$$
I claim that $(2)$ is the result of taking the polynomial
$$
\sum_{i=0}^{n+1} \binom{n+1}ix^i\tag{3}
$$
and performing the following operation $n$ times: multiply the polynomial by $x$, then differentiate. Indeed, each summand of a polynomial looks like $a_i x^i$, and when you multiply by $x$ and differentiate, the result is $(i+1)a_ix^i$. Doing this $n$ times results in $(i+1)^na_ix^i$.
But the polynomial in $(3)$ is by the binomial theorem equal to $(1+x)^{n+1}$. Note that this has a zero of order $n+1$ at $x=-1$. Multiplying by $x$ does not change this. It is a standard result that if $f(x)$ has a zero of order $k$ at $x_0$, then $f'(x)$ has a zero of order $k-1$ at $x_0$. Therefore, since $(3)$ has a zero of order $n+1$, it follows that after $n$ differentiations (and some multiplications by $x$), it will still have a zero of order $1$. In particular, $(2)$ has a zero at $x=-1$, so the expression in $(1)$ equals zero.