Sam is correct of course about $q$-hook formula. Below is a short self-contained proof not relying on such advanced combinatorics.
Denote $h_1>\ldots>h_k$ the set of hook lengths of the first column of diagram $\lambda$. Then the multiset of hooks is $\cup_{i=1}^k \{1,2,\ldots,h_i\}\setminus \{h_i-h_j:i<j\}$ and $n=\sum_i h_i-\frac{k(k-1)}2$.
Recall that $F(m)=P_m(\alpha,\beta)=\prod_{d|m,d>1}\Phi_d(\alpha,\beta)=\prod_d (\Phi_d(\alpha,\beta))^{\eta_d(m)}$, where
$\alpha,\beta=(1\pm \sqrt{5})/2$;
$P_n(x,y)=x^{n-1}+x^{n-2}y+\ldots+y^{n-1}$;
$\Phi_d$ are homogeneous cyclotomic polynomials;
$\eta_d(m)=\chi_{\mathbb{Z}}(m/d)$ (i.e., it equals to 1 if $d$ divides $m$, and to 0 otherwise).
Therefore it suffices to prove that for any fixed $d>1$ we have
$$
\sum_{m=1}^n \eta_d(m)+\sum_{i<j}\eta_d(h_i-h_j)-\sum_{i=1}^k\sum_{j=1}^{h_i}\eta_d(j)\geqslant 0.\quad (\ast)
$$
$(\ast)$ rewrites as
$$
[n/d]+|i<j:h_i\equiv h_j \pmod d|-\sum_{i=1}^k [h_i/d]\geqslant 0.\quad (\bullet)
$$
LHS of $(\bullet)$ does not change if we reduce all $h_i$'s modulo $d$ (and accordingly change $n=\sum_i h_i-\frac{k(k-1)}2$, of course), so we may suppose that $0\leqslant h_i\leqslant d-1$ for all $i$. For $a=0,1,\dots, d-1$ denote $t_a=|i:h_i=a|$. Then $(\bullet)$ rewrites as
$$
\left[\frac{-\binom{\sum_{i=0}^{d-1} t_i}2+\sum_{i=0}^{d-1} it_i}d\right]+
\sum_{i=0}^{d-1} \binom{t_i}2\geqslant 0. \quad (\star)
$$
It remains to observe that LHS of $(\star)$ equals to
$$
\left[\frac1d\sum_{i<j}\binom{t_i-t_j}2 \right].
$$
Best Answer
Multiply the whole equality by $(-1)^n$.
First of all, notice that $k\choose m_1,\dots,m_k$ is the number of ways to permute the numbers $\lambda_1,\dots,\lambda_k$, so the left-hand side equals $$ L=\sum_{\lambda_i\geq 1, \; \lambda_1+\dots+\lambda_k=n} (-1)^k\prod_{i=1}^k{n+1\choose \lambda_i}. $$ Denote $$ S(x)=\sum_{j=1}^{n+1} {n+1\choose j}x^l=(1+x)^{n+1}-1. $$ Then, due to the formula above, $L$ is the coefficient of $x^n$ in $$ 1-S(x)+S(x)^2-\dots=\frac1{1+S(x)}=\frac1{(1+x)^{n+1}} =\sum_{i=0}^\infty(-1)^i{n+i\choose i}x^i, $$ thus $L=(-1)^n{2n\choose n}$, as desired.