"Biggest" coefficients of a linear combination between vectors of zeros and ones Let $n$ be a positive integer. Denote by $B_n$ the set of $n\times(n+1)$-matrices of rank $n$ and with coefficients in $\{0,1\}$. I would like to measure how "complex" the coefficients of a linear combination of the columns of a matrix of $B_n$ can be. More precisely, I'd like to compute (or estimate the asymptotic behaviour of) 
$$
P_n := \max \left\{
\prod_{i=1}^{n+1}|\lambda_i|,\;
M\in B_n,\;
\sum_{i=1}^{n+1}\lambda_i m_{\star,i} = 0,\; 
\lambda_1,\dots,\lambda_{n+1}\in \mathbb{Z},\; \gcd(\lambda_1,\dots,\lambda_{n+1})=1 
\right\}
$$
where $m_{\star,i}$ stands for the $i$-th column vector of the matrix $M$.
$P_2=1$ and $P_3 = 2$, and for every $1\le k\le n-2$ with $\gcd(k,n-1)=1$, $P_n$ is bounded below by $(n-1)(n-(k+1))^{k}k^{n-k}$ (indeed, setting $v$ the vector whose $k$ first coefficients are 1 and $n-k$ last coefficients are 0 and $\hat e_i$ the vector whose only 0 coefficient is at line i and whose other coefficients are 1, we have the following combination: $
(n-1) v + (n-(k+1))(\hat e_1 +\dots + \hat e_k) = k(\hat e_{k+1} +\dots + \hat e_n).)
$
 A: We can show that $P_n\le (n+1)^{n(n+1)}$ for each $n\ge 2$, becuase a few months ago I proved a following lemma. 
For a natural number $n$ let $[n]$ denotes a set ${1,\dots, n}$. Given a subset $Y$ of a vector space $X$ over $\mathbb R$ by $\langle Y\rangle$ we denote the linear hull of $Y$ in $X$, that is a set of all finite sums $f_1y_1+\dots+f_ky_k$, where $f_i\in\mathbb R$ and $y_i\in Y$ for each $i$.
Lemma. Let $K$ and $N$ be positive integers,
$V=\{v_1,\dots, v_k\}\subset [0,K]^N$ be a linearly dependent over $\mathbb R$ system of vectors with
integer entries. There exist integers
$f_1,\dots, f_k$ which are not all zeroes such that $|f_i|\le (kK)^{k-1}$ for each $i$ and
$f_1v_1+\dots+f_kv_k=0$.
Proof. Let $W$ be a maximal linearly independent subset of a set $V$. Since the set $V$ is linearly
dependent, $|W|\le k-1$. For each $i\in [N]$ let
$e^i=(e^i_1,\dots,e^i_N)\in\mathbb R^N$ be $i$-th standard orth, that is $e^i_i=1$ and $e^i_j=0$ for each $j\ne i$. Let
$B_0=\{e^1,\dots,e^n\}$ be the standard basis of the linear space $\mathbb R^N$.
By [Lan, Ch. III, Theorem 2], there exists a basis $B$ of the space $\mathbb R^N$
such that $W\subset B\subset W\cup B_0$. Let $C=B_0\setminus (B\setminus W)$ and
$p_{C}:\mathbb R^N\to \langle C\rangle$ be the orthogonal
projection, that is $p_{C}(x)=\sum\{x_ie^i:x_i\in\mathbb R$, $e^i\in C\}$
for each vector $x=(x_1,\dots,x_N)\in \mathbb R^N$. Thus
$\ker p_{C}=\{x\in \mathbb R^N:p_{C}(x)=0\}=\langle B_0\setminus C\rangle=
\langle B\setminus W\rangle$. We have $\ker p_{C}\cap \langle W\rangle=\langle B\setminus
W\rangle\cap\langle W\rangle=0$, because otherwise the set $B$ is linearly dependent.
Thus the restriction $p_{C}|\langle W\rangle$ of the map $p_{C}$ on the set $\langle W\rangle$
is injective.
Put $K'=(kK)^{k-1}$. Define a map $f$ from the subset $D^k$ of points of the set $[0, K']^k$ with all
integer coordinates to $\langle W\rangle\cap \mathbb Z^N\subset \mathbb R^N$ as follows.
Let $d=(d_1,\dots,d_k)\in D^k$. Put $f(d)=p_C(dv)$, where $dv=d_1v_1+\dots d_kv_k$.
Since $d_i\in [0, K']$ and $v_i\in [0,K]^N$ for each $i\in [k]$, each coordinate of a vector
$dv$ (and, hence, of the vector $f(d)=p_C(dv)$ too) is at most $kK'K$. Since
$$|C|=|B_0\setminus (B\setminus W)|=|B_0|-|B\setminus W|=|B_0|-(|B|-|W|)=
N-(N-|W|)=|W|\le k-1,$$
$|f(Q)|\le (kK'K+1)^{k-1}$. We have $|D^k|>|f(Q)|$, because $(1+(kK)^{k-1})^{\frac 1{k-1}}>(1+(kK)^k)^{\frac 1{k}}$,
because when $a>1$ is a constant and $x>0$ a function $(1+a^x)^{\frac 1x}$ decreases.
Therefore the function $f$ is not injective. So there exist distinct elements $d=(d_1,\dots,d_k)$
and $d'=(d'_1,\dots,d'_k)$ of $D^k$ such that $p_C(dv)=f(d)=f(d')=p_C(d'v)$. Since $dv$ and $dv'$
belong to $\langle W\rangle$ and the restriction $p_{C}|\langle W\rangle$ is injective,
$dv=d'v$. It remains to put $f_i=d_i-d'_i$ for each $i\in [k]$.$\square$
Remark that for each $B_n$, the sequence $(\lambda_1,\dots,\lambda_{n+1})$ is determined up to a multiplication by $(-1)$.  The lemma implies that  $|\lambda_i|\le (n+1)^n$ for each $i$, so $P_n\le (n+1)^{n(n+1)}$.
References
[L] Serge Lange, Algebra, Addison-Wesley, 1965 (Russian translation, Moskow, Mir, 1968).
